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FRED T. HODGSON 












Modem 

Carpentry and Joinery 

Vol. II 

ADVANCED SERIES 


BEING A COMPILATION 

OF 

THE VERY BEST THINGS AND MOST MODERN AND 
PRACTICAL METHODS KNOWN IN THE ARTS 
OF CARPENTRY AND JOINERY 


PREPARED AND EDITED BY 

FRED T. HODGSON, 


Author Practical Treatise on the Uses of “The Steel Square,” 
“Modern Estimator and Contractors’ Guide,” “Common 
Sense Stair-Builder,” “Up-To-Date Hardwood 
Finisher,” “Practical Wood Carving,” &c. 


OVER 400 PRACTICAL ILLUSTRATIONS 

.------- 



> > 

> > > ^ 

CHICAGO 

FREDERICK J. DRAKE & CO., PUBLISHERS 

1906 



















LIBRARY of CONGRESS 

Two Conies Received 

SEP 6 1906 

Copyright Entry 
CLASS 4 CL XXc. No. 

(MN-7 


Copyright, 1906 
By Frederick J. Drake & CO. 
Chicago. 






PREFACE 


In a previous volume—Modern Carpentry—I made 
a fairly successful attempt to put together a series of 
plain and simple examples of working problems and 
their solutions suited to the capacity of all ordinary 
mechanics, and owing to the fact that many thousands 
of that volume have found their way into the hands of 
carpenters and joiners in the United States, Mexico 
and Canada, and to the other fact that the author, as 
well as the publishers, have received hundreds of let¬ 
ters asking for something more on the same lines, of a 
higher grade, it has been determined to make another 
or more volumes, on the same subject, hence “Modern 
Carpentry and Joinery” advanced course. 

In the present work I have drawn largely from ac¬ 
knowledged authorities and from workmen of recog¬ 
nized ability to which have been added the results of 
my awn experience and observation and my knowledge 
of the kinks and secrets of the woodworking trades. 
In a work of this kind the reader must expect to find 
something he has met in other places, and perhaps in 
other adaptations, but I think that upon careful anal¬ 
ysis ha will find the presentation of the cases somewhat 
improved, and rendered in a more simple and under¬ 
standable manner. The selections, too, I think will be 
found more suitable and more appropriate to the pres¬ 
ent-day practice than most of the matter found in re¬ 
cent technical literature on the subject. At any rate, 

5 



6 


PREFACE 


it has been my endeavor in the formation of the present 
volume to place in a handy form, instructive examples 
of the better class of work of the carpenters and join¬ 
ers’ “Art.” 

It is not intended to repeat what has already been 
published in the first volume of Modern Carpentry, un¬ 
less such repetition will be necessary to explain or 
formulate some similar matter. 

Young men are apt to think that because they have 
a fair knowledge of their trades, they know all that is 
required and if they peruse the first volume of this 
work, and have mastered its contents, they have 
reached the limit. This, of course, is a great mistake, as 
will readily be discovered by a glance over the matter 
of the present volume, and I can say right here, and 
now, that even the present volume does not by any 
means cover the whole subject, for, at least a half dozen 
other volumes could be written without touching on the 
other two already in the market, and the subject would 
not nearly be exhausted. 

It is not necessary for me to quote the authorities 
from whom I have drawn, unless the matter is of such 
importance as to demand special recognition. I may 
say, however, that in very few books written during 
the last 25 years on the subject of carpentry, there has 
been but very little presented that had not been pub¬ 
lished before in some form or other, but the descriptions 
generally, in most cases—not in all—have been im¬ 
proved more or less: and the present book does not dif¬ 
fer a great deal from most others that have preceded it, 
only that it is more up-to-date, and more suited to 20th 
century requirements. 


PREFACE 


7 


It must also be understood, that while this book goes 
out to the public under my name as author, I do not 
claim authorship, for really, it is more of a compilation 
than an original work, but, I do claim that the selec¬ 
tions and compilations made are better and more suited 
to the wants of the present-day workman than can be 
found in any other similar work published in this or 
any other country. 

FRED T. HODGSON, 
Editor and Compiler. 


Collingwood, July 1, 1906. 



PART I. 

SOLID GEOMETRY. 
INTRODUCTORY. 

In the first volume of Modern Carpentry I gave a short 
treatise on plain or Carpenters’ Geometry, which I trust 
the student has mastered, and thus prepared himself 
for a higher grade in the same science, namely, solid 
Geometry: and to this end, the following treatise has 
been selected, as being the most simple and the most 
through available. 

Solid geometry is that branch of geometry which 
treats of solids—i. e., objects of three dimensions 
(length, breadth and thickness). By means of solid 
geometry these objects can be represented on a plane 
surface, such as a sheet of paper, in such a manner that 
the dimensions of the object can be accurately mea- 
sured from the drawing by means of a rule or scale. 
The “geometrical” drawings supplied by the architect 
or engineer for the builder’s use are, with few excep¬ 
tions, problems in solid geometry, and therefore a cer¬ 
tain amount of knowledge of the subject is indispen¬ 
sable, not only to the draughtsman who prepares the 
drawings, but also to the builder or workman who has 
to interpret them. 

As the geometrical representations of objects consist 
entirely of lines and points, it follows that if projec¬ 
tions of lines and points can be accurately drawn, the 

9 


10 


MODERN CARPENTRY 


representation of objects will present no further diffi¬ 
culty. A study of lines and points, however, is some¬ 
what confusing, unless the theory of projection has first 
been grasped, and for this reason the subject will be 
introduced by a simple concrete example, such, as a 



■Vertical Projections (or Elevations) and Horizontal Projection (or Plan) of a 
Table in the Middle of an Oblong Room 

Fig. 1. 


table standing in the middle of a room. The four legs 
rest on the floor at A, B, C, and D (Fig. 1), and the 
perpendiculars let fall from the corners of the table- 
top to meet the floor at E, F, G, and H. The oblong 
EFGH represents the horizontal projection or “plan” 



































SOLID GEOMETRY 


11 


of the table-top; the small squares at A, B, C, and D 
represent the plan of the four legs, and the lines cor¬ 
recting them represent the plan of the frame work 
under the top. The large oblong J K L M is a plan of 
the room. 

The plan or horizontal projection of an object is 
therefore a representation of its horizontal dimensions 
—in other words, it is the appearance which an object 
presents when every point in it is viewed from a posi¬ 
tion vertically above that point. 

In a similar manner an elevation or vertical projec¬ 
tion of an object is a representation of its vertical di¬ 
mensions, and also, it should be added, of some of its 
horizontal dimensions,—i. e., it is the appearance which 
an object presents when every point in it is viewed from 
a position exactly level with that point, all the lines 
of sight being parallel both horizontally and vertically. 
Thus, the front elevation of the table (or the vertical 
projection of the side GH) will be as shown at 
G' IF C' D', and the vertical projection of the side JK 
of the room will be JK JK. The vertical projection of 
the end of E G of the table will be as shown E" G" A" 
C" and of the end K M of the room will be K" M" K M; 
the drawing must be turned until the line K M is hori¬ 
zontal, for these and projections to be properly seen. 

By applying the scale to the plan, we find that the 
length of the table-top is six and a half feet, the 
breadth 4 feet, and the distance of the table from each 
wall 4% feet. From the front elevation we can learn 
the height of the table, and also give its length and dis¬ 
tance from the end walls of the room. From the end 
elevation we can ascertain the breadth of the table and 


12 


MODERN CARPENTRY 


its distance from the sides of the room, as well as its 

♦ 

height. 

To make the drawing clearer, let us imagine that 
the walls of the room are of wood and hinged at the 
level of the floor. On the wall J K draw the front ele¬ 
vation of the table and then turn the wall back on its 
hinges until it is horizontal,—i. e., in the same plane 
as the floor. Proceed in a similar manner with the end 
K M, and we get the three projections of the room and 
table on one plane, as shown in the diagram. To avoid 
confusion the end elevation will not be further con¬ 
sidered at the present. 

It will be seen that the line J K represents the angle 
formed by the wall and floor,—in other words, it repre¬ 
sents the intersection of the vertical and horizontal 
‘‘planes of projection,” it is known as “the line of in¬ 
tersection,” or “the ground line.” If a line is let fall 
from G' perpendicular to J K, the two lines will meet 
at G, and they will be in the same straight line. Simil¬ 
arly, the perpendiculars H' h and h H are in the same 
straight line. Lines of this kind perpendicular to the 
planes of projections are known as “projectors,” and 
are either horizontal or vertical G' g and H' h are ver¬ 
tical projectors; G g, C c, D d, and ITh are horizontal 
projectors. 

Vertical projectors are not always parallel to one of 
the sides of the object represented, or, if parallel to one 
side, are not parallel to other sides which must be rep¬ 
resented; thus, a vertical projector or “elevation” of 
an octagonal object, if parallel to one of the sides of the 
octagon, must be oblique to the two adjacent sides. In 
Fig. 2 an octagonal table is shown. The plan must first 


SOLID GEOMETRY 


13 


be drawn, and from the principal points of the plan 
projectors must be drawn perpendicular to the vertical 
plane of the projection, until they cut the ground line, 
and from this perpendiculars must be erected to the 
height of the several parts of the table. The elevation 


B 



(Plan and Elevation of an Octagonal Table 

Fig - . 2. 


can then be completed without difficulty. The side 
B C of the table is parallel to the vertical plane of pro¬ 
jection, but the adjacent sides AB and CD are 
oblique.* 


*A distinction must be made between “perpendicular” and 
“vertical.” The former means, in geometry, a line or plane at 
right angles to another line or plane, whether these are hori¬ 
zontal, vertical or inclined; whereas a vertical line or plane 
is always at right angles to a horizontal line or plane. The 
spirit-level gives the horizontal line or plane, the plumb-rule 
gives the vertical. 































14 


MODERN CARPENTRY 


2. POINTS, LINES, AND PLANES. 

1. To determine the position and length of a given 
straight line, parallel to one of the planes of the pro¬ 
jection. 

Let GII (Fig. 3) be the given straight line. To de¬ 
termine its position (i. e., in regard to horizontal and 
vertical planes), it is only necessary to determine the 
position of any two of the extreme points. 


A 

^ B 

g 

C 

8" 0 


'g' 

E 

F 


0 3 6 9 12. in. 

L.i. i—I—i—i—L_i_i_I_i i I 



•No. 1. Perspective View. No. 2. Geometrical Projections 
Horizontal and Vertical 

Fig. 3. 

Let A B C D be a vertical plane parallel to the given 
line, and C D E F a horizontal plane. The vertical pro¬ 
jection or elevation of the line is represented by the 
line li g, and the horizontal projection or plan by the 
point g', the various projection being shown by dotted 
lines. The given line is proved to be vertical, because 

















SOLID GEOMETRY 


15 


its horizontal projection is a point; its length as mea¬ 
sured by the scale, is 6 inches; its height (g g") above 
the line of intersection C D is 4 inches; and its hori¬ 
zontal distance (g' g") from the same line is 8 inches. 

If the illustrations are turned so that C D E F be¬ 
come a vertical plane, and A B C D the horizontal plane 
then GII will be horizontal line, because one of its ver¬ 
tical projections is a point. Other vertical projections 
of the line can be made,—as, for example, a side ele¬ 
vation,—in which the projection will appear as a line 
and not a point, but a line must be horizontal if any 
vertical projection of it is a point. 



Fig. 4. 

Let the given line GH (Fig. 4) be parallel to the 
vertical plane, but inclined to the horizontal plane. 
Then g h will be a vertical projection, and g' h' its hori¬ 
zontal projection or plan. By producing h g till it cuts 
C D at c, it will be found that the given line is inclined 
at an angle of 60° to the horizontal plane; its length, 
as measured by the scale along the vertical projection 
































16 


MODERN CARPENTRY 


gh, is 8 inches; the height of G above the horizontal 
plane (measured at g g") is 3 inches, and the height of 
H (measured at hh") is 9 % inches. 

II. To determine the length of a given straight line, 
which is oblique to both planes of projection. 



Let a b (Fig. 5) be the horizontal projection and a' b' 
the given vertical projection of the given line. Draw 
the projection A B on a plane parallel to ab in the 
manner shown. From this projection the length of the 
given line will be found (b} r applying the scale) to be 
10 inches. The height C B is of course equal to the 
height c' b', and the horizontal measurement A C is 
equal to the horizontal measurement A C is equal to the 









SOLID GEOMETRY 


n 


horizontal projection a b, and the angle A C B is a right 
angle. It follows therefore that, if from b the line b d 
is drawn perpendicular to a b and equal to the height 
c' b', the line joining a d will be the length of the given 
i line. To avoid drawing the horizontal line a' c' the 
1 height of a' and b' above x y are usually set up from a 
and b as at a" and b", the line a" b" is the length of 
the given line.* 

III. The projections of a right line being given, to 
find the points wherein the prolongation of that line 
would meet the planes of projection. 




-I. Perspective View; II. Vertical and Horizontal Projections 

Fig. 6. 

Let a b and a' b' (Fig. 6, II) be the given projection 
of the line A B. In the perspective representation of 
the problem it is seen that A B, if prolonged, cuts the 
horizontal plane in c, and the vertical plane in d, and 
the projections of the prolongation becomes c e am] 
fd. Hence, if a b, a'b' (Fig. 6, II) be the projections 

*The application of this problem to hips is obvious. Sup¬ 
pose that a & is the plan of a hip-rafter, and a' b' an eleva* 
tion, the length of the rafter will be equal to a d or a b. 















18 


MODERN CARPENTRY 


of A B, the solution of the problem is obtained by pro¬ 
ducing these lines to meet the common intersection of 
the planes in f and e, and on these points raising the 
perpendiculars fc and ed, meeting a b produced in c 
and a' b' produced in d; c and d are the points sought.* 

IV. If two lines intersect each other in space, to 
find from their given projections the angles which they 
make with each other. 

Let ab, c d, and a' b' c'd' (Fig. 7) be the projections 
of the lines. Draw the projectors e' f' f' e', perpendicu¬ 
lar to the line of intersection a' c', and produce it in¬ 
definitely towards E"; from e draw indefinitely, e E' 
perpendicular to the line e' f, and make e E' equal to f" 
e, and draw f E'. From f as a centre describe the arc 
E"gE", meeting ef produced in E", and join a E", c 
E". The angle a E" c is the angle sought. This problem 
is little more than a development of problem II. If we 
consider e f, e' f', as horizontal and vertical projections 
of an imaginary line lying in the same plane as a e and 
c e, we find the length of this line by problem II to be 
f E'; in other words f E' is the true altitude of the tri¬ 
angle a e c, a' e' c'. Construct a triangle on the base a e 
with an altitude f E' equal to f E', and the problem is 
solved. The practical application of this problem will 
be understood if we imagine a e c to be the plan and 
a' e' c' the elevation of a hipped roof; f E' gives us the 

*The points d and c are known respectively as the verti¬ 
cal and horizontal “traces” of the line a b, the trace of a 
straight line on a plane being the point in which the straight 
line, produced if necessary, meets or intersects the plane. A 
horizontal line cannot therefore have a horizontal trace, as it 
cannot possibly, even if produced, meet or intersect the hori¬ 
zontal plane; for a similar reason, a vertical line cannot have 
a vertical trace. 



SOLID GEOMETRY 19 

length and slope of the longest common rafter or spar, 
and a E" C is a true representation of the whole hip, i. 
e., on a plane parallel to the slope of the top. It will be 
observed that the two projections (e and e') of the 



point of intersection of the two lines are in a right line 
perpendicular to the line of intersection of the plains 
of projection. Hence this corollary.— The projections 
of the point of intersections of two lines which cut 







20 


MODERN CARPENTRY 


each other in space are in the same right line perpen¬ 
dicular to the common intersection of the planes of 

projection. This is further illustrated by the next 
problem. 




Fig - . S. 


V. To determine from the projection of two lines 
which intersect each other in the projections, whether 
the lines cut each other in space or not. 

Let a b, c d, a b', c d' (Fig. 8, I) be the projections of 
the lines. It might be supposed, as their projections in- 










SOLID GEOMETRY 


21 


tersect each other that the lines themselves intersect 
each other in space, but on applying the corollary of 
the preceding problem, it is found that the intersec¬ 
tions are not in the same perpendicular to the line of 
intersection a c of the planes of projection. This is rep¬ 
resented in perspective in Fig. 8, II. We there see that 
the original lines a B c D do not cut each other, although 
their projections a b, c d, a b', c d', do so. From the point 
of intersection e raise a perpendicular to the horizontal 
plane, and it will cut the original line c D in E, and this 
point therefore belongs to the line c D, but c belongs 
equally to a B. As the perpendicular raised on e passes 
through E on the line c D, and through E on the line 
a B, these points E E' cannot be the intersection of the 
two lines, since they do not touch; and it is also the 
same in regard to f f. Hence, when two right lines do 
not cut each other in space, the intersections of their 
projections are not in the same right line perpendicular 
to the common intersection of the planes of projection. 

VI. To find the angle made by a plane with the 
horizontal plane of projection. 

Let a b and a c, Fig. 9, be the horizontal and vertical 
traces of the given plane, i. e., the lines on which the 
given plane would, if produced, cut the horizontal and 
vertical planes of projection. Take any convenient point 
d in a b, and from it draw d e perpendicular to a b, 
and cutting the line of intersection x y in e, from e 
draw e d perpendicular to x y and cutting a c in d'. The 
angle made by the given plane with the horizontal plane 
of projection is such that, with a base d e, it has a ver¬ 
tical height e d\ Draw such an angle on the vertical 


22 


MODERN CARPENTRY 


plane of projection by setting off from e the distance 
e d" equal to e d, and joining d'd". The angle d'd" e 
is the angle required.* 



VII. The traces of a plane and the projections of a 
point being given, to draw through the point a plane 
parallel to the given plane. 

In the prospective representation (Figs. 10 and 11) 
suppose the problem solved, and let B C be the given 

*The angle made by a plane with the vertical plane of pro¬ 
jection can be found in a similar manner. If we imagine the 
part above xy in fig. nine to be the horizontal projection and 
the part below xy to be the vertical projection—in other 
words, if fig nine is turned upside down —a b becomes the ver¬ 
tical trace and a c the horizontal trace, and the angle cl' 
xl" e is the angle made by the given plane with the vertical 
plane of projection. 









SOLID GEOMETRY 


23 


plane, and A C, A B its vertical and horizontal traces, 
and E F a plane parallel to the given plane, and G F, 
G E its traces. Through any point D, taken at pleasure 



I. Vertical and Horizontal Projections. II. Persoective View 

Fig. 10. 


on the plane E F, draw the vertical plane IIJ, the hori¬ 
zontal trace of which, I H, is parallel to G E. The plane 
H J cuts the plane E F in the line k 1', and its vertical 
trace G F in T. The horizontal projection of k' T is H I, 














24 


MODERN CARPENTRY 


and its vertical projection k' T; and as the point D is in 
k 1', its horizontal and vertical projections will be d 
and d'. Therefore, if through d be traced a line d 1, 
parallel to A B, that line will be the horizontal projec¬ 
tion of a vertical plane passing through the original 
point D; and if an 1 be drawn the indefinite perpen¬ 
dicular L" T, and through d', the vertical projection of 
the given point, be drawn the horizontal line d' 1', cut¬ 
ting the perpendicular in T, then the line F G drawn 
through 1 parallel to A C, will be the vertical trace of 
the plane required; and the line G E drawn parallel 



to A B, its horizontal trace. Hence, all planes parallel 
to each other have their projections parallel, and re¬ 
ciprocally. In solving the problem, let AB, A C (Figs. 
10, I) be the traces of the given plane, and d d the 
projections of the given point. Through d draw d 1 
parallel to A B, and from I draw 11' perpendicular 
to A II. Join d d' and through d' draw d' 1' parallel 
to the line of intersection AII. Then F T G drawn 
parallel to A C, and G E parallel to A B, are the 
traces of the required plane, 






SOLID GEOMETRY 


25 


VII. The traces A B, B C, and A D, D C, of two 
planes which cut each other being given, to find the 
projections of their intersections. 

The planes intersect each other in the straight line 
A C (Fig. II, 11), of which the points A and C are the 
traces, since in these points this line intersects the 
planes of projection. To find these projections, it is 
only necessary to let fall on the line of intersection in 
Fig. II, 1, the perpendiculars A a, C c, from the points 
A and C, and join A c, C a. Ac will be the horizontal 
projection, and C a the vertical projection of C A, which 
is the line of intersection or arris of the planes. 

XI. The traces of two intersecting planes being 
given, to find the angle which the planes make be¬ 
tween them. 

The angle formed by two planes is measured by that 
of two lines drawn from the same point in their inter¬ 
section (one along each of the planes), perpendicular 
to the line formed by the intersection. This will be bet¬ 
ter understood by drawing a straight line across the 
crease in a double sheet of note-paper at right angles 
to the crease; if the two leaves of the paper are then 
partly closed so as to form an angle, we have an angle 
formed by two planes, and this angle is the same as that 
formed by two lines which have been drawn perpen¬ 
dicular to the line of intersection of the two planes. 
These lines in effect determine a third plane perpendicu¬ 
lar to the arris. If, therefore, the two planes are cut 
by a third plane at right angles to their intersection 
the solution of the problem is obtained. 


26 


MODERN CARPENTRY 




l. Vertical and Horizontal Projections. II. Perspective Vj$W 

Fig. 12. 













SOLID GEOMETRY 


27 


On the arris A C (Pig. 12, 11) take at pleasure any 
point E, and suppose a plane passing through that point 
cutting the two given planes perpendicular to the arris. 
There results from the section a triangle D E F, inclined 
to the horizontal plane, and the angle of which, D E F, 
is the measure of the inclination of the two planes. The 
horizontal projection of that triangle is the triangle 
D e F, the base of which, F D, is perpendicular to A c 
(the horizontal projection of the arris A C), and cuts it 
in the point g, and the line E g is perpendicular to D F. 
The line g E is necessary perpendicular to the arris A C, 
as it is in the plane D E F, and its horizontal projection 
is g e. Now, suppose the triangle D E F turned on D F 
as an axis, and laid horizontally, its summit will then be 
at E", and D E" F is the angle sought. The perpendic¬ 
ular g E is also in the vertical triangle A C c, of which 
the arris is the hypothenuse and the sides Ac, C C are 
the projections. This description introduces the so¬ 
lution of the problem. 

Through any point g (Fig. 12, 1) on the line A c, the 
horizontal projection of the arris or line of intersection 
of the two planes, draw FD perpendicular to Ac; on 
A c (which for the moment must be considered as a 
‘Dine of intersection” or “ground line” for a second 
vertical projection) describe the vertical projection of 
the arris by drawing the perpendicular c C, and then 
joining A C'. A C gives the true length and inclination 
of the arris. From g draw g E' perpendicular to A C', 
and meeting it in E'; from E' let fall a perpendicular 
E e' or A c, meeting Ac in e. F e D is the horizontal 
projection of the triangle FED (see No. 11) and from 
this the vertical projection f' e' D can be drawn as 


28 


MODERN CARPENTRY 


shown. We have now obtained the vertical and hori¬ 
zontal projections of two intersecting lines, namely, 
F e, e D, and f' e', e' D, and by problem (4) the triangle 
which they make with each other can be found. It will 
be seen that g E' is the true altitude of the triangle 
f' e' D, as c E is equal to e' e. Set off therefore from g 



I. Horizontal Projection. II. Vertical Projection on Plane AD. 
III. Vertical Projection on Plane CD 

Fig. 13. 


towards A a distance g E equal to g E, and join F E, 
E" D; FE" D is the angle made by the two intersect¬ 
ing planes. 














SOLID GEOMETRY 


29 



‘I. Vertical ami Horizontal Projections. II. Perspective View 


Fig. 14. 

















30 MODERN CARPENTRY 

X. Through a given point to draw a perpendicular 
to a given plane. 

Let a and a' (Fig. 14, II.) be the projections of the 
given point, and BC, CD the horizontal and vertical 
traces of the given plane. Suppose the problem solved, 
and that A E is the perpendicular drawn through the 
A to the plane B D, and that its intersection with the 
plane is the point E. Suppose also a vertical plane 
a F to pass through A E', this plane would cut B D in 
the line g D, and its horizontal trace a h would be 
perpendicular to the trace B C. In the same way 
a' e', the vertical projection of A E, would be perpen¬ 
dicular to C D, the vertical trace of the plane B D. 
Thus we find that if a line a h is drawn from a, per¬ 
pendicular to B C, it will be the horizontal trace of 
the plane in which lies the required perpendicular 
A E, and h F will be the vertical trace of the same 
plane. From a', draw upon C D an indefinite perpen¬ 
dicular, and that line will contain the vertical pro¬ 
jection of A E, as a li contains its horizontal pro¬ 
jection. To find the point of intersection of the line 
A E with the given plane, construct the vertical pro¬ 
jection g D of the line of intersection of the two 
planes, and the point of intersection of that line with 
the right line drawn through a', will be the point 
sought. If from that point a perpendicular is let 
fall on a h, the point e will be the horizontal projec¬ 
tion of the point of intersection E. 

In Fig. 15, 1, let B C, CD be the traces of the 
given plane, and a, a' the projections of the given 
point. From the point a, draw a h perpendicular to 
BC; ah will be the horizontal projection of a plane 


SOLID GEOMETRY 


31 


passing vertically through a, and cutting the given 
plane. On ah as ‘‘ground line” draw a vertical pro¬ 
jection as follows: Fl’om a, draw a A perpendicular 
to a g, and make it equal to a" a'; from li draw li D' 
perpendicular to h a, and make h D' equal to h D; 
draw g D, which will be the section of the given plane 
by a vertical D g, and the angle h g D' will be the 



I. Horizontal Projection. II. Vertical Projection on Plane parallel to 
ha". III. Vertical Projection on Plane parallel to A a 

FI g. 15. 


measure of the inclination of the given plane with the 
horizontal plane; there is now to be drawn, perpen¬ 
dicular to this line, a line A E through A, which will 
be the vertical projection on a h of the line required. 
From the point of intersection E let fall upon aha 














MODERN CARPENTRY 


32 


perpendicular, which will give e as the horizontal pro¬ 
jection of E. Therefore a e is the horizontal projection 
of the required perpendicular, and a' e' its vertical 
projection on the original “ground line” ha". It 
follows from this problem that— Where a right line in 
space is perpendicular to a plane, the projections of 
that line are respectively perpendicular to the traces 
of the plane. 

XI. Through a given point to draw a plane perpen¬ 
dicular to a given right line. 

Let a, a' (Fig. 16, 1) be the projections of the given 
point A, and b c, b' c' the projections of the given 
line B C. 

The foregoing problem has shown that the traces of 
the plane sought must be perpendicular to the pro¬ 
jections of the line, and the solution of the problem 
consists in making to pass through A, a vertical plane 
Af (Fig. 16, 11), the horizontal projection of which 
will be perpendicular to b c. 

Through a (Fig. 16, 1) draw the projection af 
perpendicular to b c. From f raise upon K L the in¬ 
definite perpendicular f f, which will be the vertical 
trace of the plane a f f perpendicular to the horizontal 
plane, and passing through the original point A (in 
No. 11). Then draw through a in the vertical pro¬ 
jection a horizontal line, cutting f f in f, which point 
should be in the trace of the plane sought; and as 
that plane must be perpendicular to the vertical pro¬ 
jection of the given right line draw through f a per¬ 
pendicular to b c, and produce it to cut K L in G. 
This point G is in the horizontal trace of the plane 
sought. All that remains therefore, is from G to draw 


SOLID GEOMETRY 33 






















34 


MODERN CARPENTRY 


G D perpendicular to b c. If the projections of the* 
straight line are required, proceed as in the previous 
problem, and as shown by the dotted lines; the plane 
will cut the given line at k in the horizontal projec¬ 
tion, and at k in the vertical projection. * 

XII. A right line being given in projection, and 
also the traces of a given plane, to find the angle which 
the line makes with the plane. 

Let AB (Fig. 17, 11) be the original right line inter¬ 
secting the plane C E in the point B. If a vertical 
plane a B pass through the right line, it will cut the 
plane C E in the line f B, and the horizontal plane in 
the line a b. As the plane a B is in this case parallel 
to the vertical plane of projection,its projection on 
that plane will be a quadrilateral figure a b of the 
same dimensions, and fB contained in the rectangle 
will have for its vertical projection a right line D b, 
which will be equal and similar to f B. Hence the 
two angles a b D, ABf, being equal, will equally be 
the measure of the angle of inclination of the right 


♦The diagram will be less confusing if the projection on a h 
is drawn separately, as in Fig. 16, where I is the horizontal 
projection or plan, II the vertical projection or elevation on 
a plane parallel to ha", and III the vertical projection or ele¬ 
vation on a plane parallel to ha. To draw No. Ill draw first 
the ground-line h a equal to h a on No. I, and mark on it the 
point g; from h draw the vertical h d' equal to h n, and from 
a draw the vertical a a equal to a" a'; join d' g and a h, and 
from the point of intersection e let fall a perpendicular on h a, 
cutting it in e. a e is the actual length and inclination of the 
required line, and a e its horizontal length. Transfer the 
length a e to No. I, and from e draw e e' perpendicular to h 
a", and cutting o' h in e'. If the drawing has been correctly 
made, a line from e parallel to the ground-line h a" will also 
intersect o' h in e\ 



SOLID GEOMETRY 


35 


line A B to the plane C E. Thus the angle a b D (Fig. 
17, 1), is the angle sought. 

This case presents no difficulty; but when the line 
is in a plane which is not parallel to the plane of 




projection, the problem is more difficult; as, however, 
the second case is not of much practical value, it will 
not be considered. 














36 


MODERN CARPENTRY 


3. STRAIGHT-SIDED SOLIDS. 

XIII. Given the horizontal projection of a regular 
tetrahedron, to find its vertical projection. 

Let ABCd (Fig. 18) be the given projection of 
the tetrahedron, which has one of its faces coincident 
with the ground-line. From d draw an indefinite line 


tf 



Fig. 18. 

dD perpendicular to d C, and make CD equal to CB, 
C A, or A B; d D will be the height sought, which is 
carried to the vertical projection from c to d. Join 
A d and B d, and the vertical projection is complete. 
This problem might be solved in other ways. 







SOLID GEOMETRY 


37 


XIV. A point being given in one of the projections 
of a tetrahedron, to find the point on the other pro¬ 
jection. 

Let e be the point given in the horizontal projec¬ 
tion (Fig. 18). It may first be considered as situated 
in the plane C B d, which is inclined to the horizontal 
plane, and of which the vertical projection is the tri¬ 
angle c B d. According to the general method, the 
vertical projection of the given point is to be found 
somewhere in a perpendicular raised on its horizontal 
projection e. If through d and the point e be drawn 
a line produced to the base of the triangle in f, the 
point e will be on that line, and its vertical projection 
will be on the vertical projection of that line fed, 
at the intersection of it with the perpendicular raised 
on e. If through e be drawn a straight line g h, par¬ 
allel to C B, this will be a horizontal line, whose ex¬ 
tremity h will be on B d. The vertical projection of 
d B is d B; therefore, by raising on li a perpendicular 
to A B, there will be obtained li, the extremity of a 
horizontal line represented by h g in the horizontal 
plane. If through h is drawn a horizontal line li g 
this line will cut the vertical line raised on e in e, 
the point sought. If the point had been given in g 
on the arris c d, the projection could not be found in 
the first manner; but it could be found in the second 
manner, by drawing through g, a line parallel to C B, 
and prolonging the horizontal line drawn through h, 
to the arris c d, which it would cut in g, the point 
sought. The point can also be found by laying down 
the right-angled triangle C d D (which is the devel¬ 
opment of the triangle formed by the horizontal pro- 


38 


MODERN CARPENTRY 


jection of the arris C d, the height of the solid, and 
the length of the arris as a hypotenuse), and by draw¬ 
ing through g the line g G perpendicular to C d, to in¬ 
tersect the hypotenuse in G, and carrying the height 
gG from c to g in the vertical projection. One or 
other of these means can be employed according to 
circumstances. If the point had been given in the 
vertical instead of the horizontal projection, the same 
operations inverted would require to be used. 


d' 



XV. Given a tetrahedron, and the trace of a plane 
(perpendicular to one of the planes of projection) cut¬ 
ting it, by which it is truncated, to find the projection 
of the section. 

First when the intersecting plane is perpendicular 
to the horizontal plane (Fig. 19), the plane cuts the 








SOLID GEOMETRY 


39 


base in two points e f, of which the vertical projec¬ 
tions are e and f; and the arris B d is cut in g, the 
vertical projection of which can readily be found in 
any of the ways detailed in the last problem. Having 
found g join egfg and the triangle egf is the pro¬ 
jection of the intersection sought. 


d' 



When the intersecting plane is perpendicular to the 
vertical plane, as e f in Pig. 20, the horizontal projec¬ 
tions of the three points egf have to be found. The 
point g in this case may be obtained in several ways. 
First by drawing Ggh through g, then through h 
drawing a perpendicular to the base, produced to the 
arris at h, in the horizontal projection, and then draw¬ 
ing h g parallel to C B, cutting the arris B d in g, 
which is the point required. Second, on d B, the hor- 










40 


MODERN CARPENTRY 


izontal projection of the arris, construct a triangle 
d D B, d B being the altitude of the tetrahedron, and 
B D the arris, and transfer this triangle to the ver¬ 
tical projection at d d B. From g draw the horizontal 
line cutting B d in G; g G is the horizontal distance 
of the required point in the arris, from the vertical 
axis of the tetrahedron as d is horizontal projection 
of the vertical axis, and d B the horizontal projection 
of the arris, it follows that the length gG, trans¬ 
ferred to d g, will give the required point g. The 
points e and f are found by drawing lines from e and 
f perpendicular to the ground-line, and producing 
them till they meet the horizontal projections of the 
arrises in e and f. The triangle e f g is the horizontal 
projection of the section made by the plane e f. 

XVI. The projections of a tetrahedron being given, 
to find its projections when inclined to the horizontal 
plane in any degree. 

Let ABCd (Fig. 21) be the horizontal projection 
of a tetrahedron, with one of its sides coincident with 
the horizontal plane, and edB its vertical projection; 
it is required to find its projections when turned round 
the arris A B as an axis. The base of the pyramid 
being a horizontal plane, its vertical projection is the 
right line c B. If this line is raised to c by turning 
on B, the horizontal projection will be A c2 B. When 
the point c, by the raising of B c, describes the arc 
c c, the point d will have moved to d, and the per¬ 
pendicular let fall from that point on the horizontal 
plane will give d3, the horizontal projection of the 
extremity of the arris C d; for as the summit d moves 
in the same plane as C, parallel to the vertical plane 


SOLID GEOMETRY 


41 


of projection the projection of the summit will evi¬ 
dently be in the prolongation of the arris C d, which 
is the horizontal projection or trace of that plane. 
The process therefore, is very simple and is as fol¬ 
lows: Construct at the point B the angle required, 
cBc, and make the triangle c d B equal to cdB; 



A 

Fig. 21. 


from d let fall a perpendicular cutting the prolonga¬ 
tion of the arris C d in d3, and from c a perpendicular 
cutting the same line m c2; join B c2, A c2, B d3, A d3. 

The following is a more general solution of the 
problem: Let ABCd (Fig. 22), be the horizontal 
projection of a pyramid resting with one of its sides 
on the horizontal plane, and let it be required to 
raise, by its angle C, the pyramid by turning around 
the arris A B, until its base makes with the horizontal 








42 


MODERN CARPENTRY 


plane any required angle, as 50°. Conceive the right 
line C e turning round e, and still continuing to be 
perpendicular to A B, until it is raised to the required 
angle, as at e C. If a perpendicular be now let fall 
from C, it will give the point C as the horizontal pro¬ 
jection of the angle C in its new position. Conceive 

d 



a vertical plane to pass through the line C e. This 
plane will necessarily contain the required angle. Sup¬ 
pose, now we lay this plane down in the horizontal 
projection thus: Draw from e the line e C, making 






SOLID GEOMETRY 


43 


with e C an angle of 50°, and from e with the radius 
e c describe an arc cutting it in C. From C let fall 
on C e, a perpendicular on the point C, which will 
then be the horizontal projection of C in its raised 
position. On C e draw the profile of the tetrahedron 
C D e inclined to the horizontal plane. From D let 
fall a perpendicular on C e produced, and it will give 
d as the horizontal projection of the summit of the 
pyramid in its inclined position. Join A d, B d, A c, 
B c to complete the figure. The vertical projection of 
the tetrahedron in its original position is shown by a d 
b, and in its raised position by a, c2, d2, b, the points 
c2, and d2 being found by making the perpendiculars 
c3 c2 and d3 d2 equal to C C and d D respectively. 


H 



XVII. To construct a vertical and horizontal pro¬ 
jections of a cube, the axis 1 of which are perpendic¬ 
ular to the horizontal plane. 

If an arris of the cube is given, it is easy to find its 
axis, as this is the hypotenuse of a right-angled tri¬ 
angle, the shortest side of which is the length of an 





44 


MODERN CARPENTRY 


arris, and the longest the diagonal of a side. Conceive 
the cube cut by a vertical plane passing through its 
diagonals E G, A C (Fig. 23), the section will be the 
rectangle A E G C. Divide this into two equal right- 
angled triangles, by the diagonal E C. If in the upper 
and lower faces of the cube, we draw the diagonals 
FII, B D, they will cut the former diagonals in the 
points f and b. Now, as the lines b B, b D, f F, f II, 
are perpendicular to the rectangular plane A E G C, 
f b may be considered as the vertical projection of 
B F and D H, and from this consideration we may 
solve the problem. 



Let A E (Fig. 24) lie the arris of any cube. The 
letters here refer to the same parts as those of the 
preceding diagram (Fig. 23). Through A draw an 
indefinite line, AC, perpendicular to AE. Set off on 







SOLID GEOMETRY 


45 


this line, from A to C, the diagonal of the square of 
AE, and join E C, which is then the axis of the cube. 
Draw the lines E, G, G G, parallel respectively to A C 
and A E, and the resulting rectangle, A E G C, is the 
section of a cube on the line of the diagonal of one of 
its faces. Divide the rectangle into two equal parts 
by the line b f, which is the vertical projection of 
the lines B F, D H (Fig. 23), and we obtain, in the 
figure thus completed, the vertical projection of the 
cube, as acbd (Fig. 25). 

Through C (Fig. 24), the extremity of the diagonal 
E C draw y z perpendicular to it, and let this line 
represent the common section or ground-line of the 
two planes of projection. Then let us find the hor¬ 
izontal projection of a cube of which AEGC is the 
vertical projection. In the vertical projection the 
axis E C is perpendicular to y z, and consequently, 
to the horizontal plane of projection, and we have 
fcl ? height above this plane of each of the points which 
terminate the angles. Let fall from each of these 
points perpendiculars to the horizontal plane, the 
projections of the points will be found on these per¬ 
pendiculars. The horizontal projection of the axis E C 
will be a point on its prolongation, as c. This point 
might have been named e with equal correctness, as it 
is the horizontal projection of both the extremeties of 
the axis, C and E. Through c draw a line parallel 
to y z, and find on it the projections of the points A 
and G, by continuing the perpendiculars A a, G g, to 
a and g. We have now to find the projections of the 
points bf (representing D B FII, Fig. 23,), which will 
be somewhere on the perpendiculars b b, f f, let fall 


46 


MODERN CARPENTRY 


from them. We have seen in Fig. 23 that BF, DH 
are distant from b f by an extent equal half the diag¬ 
onal of the square face of the cube. Set off, therefore, 
on the perpendiculars b b and f f, from o and m, the 
distance. A b in d, b, and f, f and join da, a b, b f, 
f g, g f, to complete the hexagon which is the horiz¬ 
ontal projection of the cube. Join f e, f e, and a c, to 
give the arrises of the upper half of the cube. The 
dotted lines d c, b c, g c, show the arrises of the lower 
side. Knowing the heights of the points in these ver¬ 
tical projections, it is easy to construct a vertical pro¬ 
jection on any line whatever, as that on R. S below. 




XVIII. To construct me projections of a regular 
octahedron, when one of its axis is perpendicular to 
either plane of projection. 

Describe a circle (Fig. 26), and divide it into four 
equal parts by the diameters, and draw the lines a d, 
cl b, be, c a; a figure is produced which serves for 
either the vertical or the horizontal projection of the 
octahedron, when one of its axis is perpendicular to 
either plane. 












Solid geometry 


47 


XIX. One of the faces of an octahedron being given, 
coincident with the horizontal plane of projection, to 
construct the projections of the solid. 

Let the triangle ABC (Fig. 27) be the given face. 
If A be considered to be the summit of one of the two 
pyramids which compose the solid, B C will be one of 
the sides of the square base, k C B i. The base makes 
with the horizontal plane an angle, which is easily 
found. Let fall from A a a perpendicular on B c, cut¬ 
ting it in d, with the length B c as a radius, and from 



d as a centre, describe the indefinite arc e f. The per¬ 
pendicular A d will be the height of each of the faces, 
and, consequently, of that which, turning on A, should 
meet the side of the base which has already turned on d. 
Make this height turn on A, describing from that point 
as a centre, with the radius A d, an indefinite arc, cut¬ 
ting the first arc in G, the point of meeting of one of 
the faces with the square base; draw the line G A, G d: 
the first is the profile or inclination of one of the faces 
on the given face ABC according to the angle d A G; 




48 


MODERN CARPENTRY 


the second, d G, is the inclination of the square base, 
which separates the two pyramids in the angle A d G. 
The face adjacent to the side B C is found in the same 
manner. Through G, draw the horizontal line GII 
equal to the perpendicular A d. This line will be the 
profile of the superior face. Draw d II, which is the 
profile of the face adjacent to B C. From H let fall a 
perpendicular on A d produced, which gives the point 
h for the horizontal projection of IT, or the summit of 
the superior triangle parallel to the first, draw h i par¬ 
allel to C A, h k parallel to A B, A k and B h per¬ 
pendicular to A B, and join k C, C h, B i, and A i and 
the horizontal projection is complete. From the heights 
we have thus obtained we can now draw the vertical 
projection shown in No. II, in which the parts have the 
same letters of reference. 

The finding of the horizontal projection may be 
abridged by constructing a hexagon and inscribing in 
it the two triangles A C B, hi k. 

XX. Given in the horizontal plane the projection of 
one of the faces of a dodecahedron, to construct its 
projections. 

The dodecahedron is a twelve-sided solid, all the sides 
being regular and equal pentagons. It is necessary, in 
order to construct the projection, to discover the in¬ 
clination of the faces among themselves. Let the penta¬ 
gon ABODE (Fig. 28) be the side on which the body 
is supposed to be seated on the plane. Conceive two 
other faces, E F G II D and D I K L C, also in the hori¬ 
zontal plane, and then raised by being turned on their 
bases, E D, D C. By their movement they will describe 
in space arcs of circles, which will terminate by the 
meeting of the sides D II, D I. 


SOLID GEOMETRY 


49 


To find the inclination of these two faces. —From the 
points 1 and H let fall perpendiculars on their bases 
produced. If each of these pentagons were raised ver¬ 
tically on its base, the horizontal projection of H and 1 
would be respectively in z z; but as both are raised 



together, the angles II and I would meet in space above 
h where the perpendiculars intersect therefore, h will 
be the horizontal projection of the point of meeting of 
the angles. To find the horizontal projection of K, pro¬ 
long indefinitely z I, and set off from z on z I the length 












50 


MODERN CARPENTRY 


x K in k; from z as centre with radius z I, describe 
the arc 11 cutting the perpendicular h I in I; join z I; 
then from z as centre, with the radius z k, describe an 
arc cutting z I produced in the point K, from which 
let'fall on z k a perpendicular K k, and produce it to 
k in x K. If, now, the right-angled triangle z k K, were 
raised on its base, k would be the projection of K. 
Conceive now the pentagon C D I K L turned round on 
C D, until it makes an angle equal to k z K with the 
horizontal plane, the summit K will then be raised 
above k by the height k K, and will have for its hori¬ 
zontal projection the point k. In completing the figure 
practically;—from the centre o, describe two concentric 
circles passing through points h D. Draw the lines 
h D, h k, and carry the last round the circumference in 
mnoprstu: through each of these lines draw radi¬ 
ally the lines m C, o B, r A, t E, and these lines will be 
the arrises analogous to h D. This being done, the in¬ 
ferior half of the solid is projected. By reason of the 
regularity of the figure, it is easy to see that the six 
other faces will be similar in those already drawn, only 
that although the superior pentagon will have its angles 
on the same circumference as the inferior pentagon 
the angles of the one will be in the middle of the faces 
of the other. Therefore, to describe the superior half; 
—through the angles n p s v k, draw the radial lines 
n 1, p 2, s 3, v 4, k 5, and join them by the straight lines 
1 2, 2 3, 3 4, 4 5, and 5 1. 

To obtain the length of the axis of the solid, observe 
that the point k is elevated above the horizontal plane 
by the height k K: carry that height to k K: the point 
r, analogous to h, is raised the same height as that 


SOLID GEOMETRY 51 

point, that is to say h i, which is to be carried from r 
to R; and the line R K is the length sought. As this 
axis should pass through the centre of the body, if a 
vertical projection of the axis in 0, and therefore 0 o 
is the half of the height of the solid vertically. By 
doubting this height, and drawing a horizontal line to 
cut the vertical lines of the angles of the superior face 
is obtained, as in the upper portion of (Fig. 28), in 
which the same letters refer to the same parts. 

XXI. One of the faces of a dodecahedron being 
given, to construct the projections of the solid, so that 
its axis may be perpendicular to the horizontal plane. 

Let ABEDC (Fig 29, 1) be the given face. The 
solid angles of the dodecahedron are each formed by 
the meeting of three pentagonal planes. If there be 
conceived a plane B C passing through the extremities 
of the arrises of the solid angle A, the result of the 
section would be a triangular pyramid, the sides of 
whose base would be equal to one of the diagonals of 
the face, such as B C. An equilateral triangle b c f 
(Fig. 29, 11) will represent the base of that pyramid 
inverted, that is, with its summit resting on the hori¬ 
zontal projection, it is required to find the height of that 
pyramid, or which is the same thing, that one of the 
three points of its base b c f, for as they are all equally 
elevated, the height of one of them gives the others. 
There is necessarily a proportion between the triangle 
Abe (No. 11) and ABC (No. 1), since the first is the 
horizontal projection of the second. A g is the hori¬ 
zontal proection of A G; but A G is a part of A H, and 
the projection of that line is required for one of the faces 
of the solid; therefore as A G: A g:: A H: x. In other 


52 


MODERN CARPENTRY 

words, the length, the length of x may be obtained by 
drawing a fourth proportion of the three lines it will 
be found to be equal to A li; or it may be obtained 
graphically thus:—Raise on A g at g an indefinite per¬ 
pendicular, take the length A G (No. 1) and carry it 


a" 



from A to G (No. 11); g is a point in the assumed 
pyramidal base b c f. Since A G is a portion of A H, 
A G will be so also. Produce A G, therefore, to IT, mak¬ 
ing All equal to AII (No. 1) and from II let fall a 
























SOLID GEOMETRY 


53 


perpendicular on A g produced, which gives h the point 
sought. Produce LI h, and carry on it the length H D 
or II E from h to d and h to e; draw the lines c d, b e, 
and the horizontal projection of one of the faces is ob¬ 
tained inclined to the horizontal plane, in the angle 
H A h. As the other two inferior faces are similar to 
the one found, the three faces should be found on the 
circumference of a circle traced from A as centre, and 
with A d or A e as a radius. Join f A, prolong An, 
A o, perpendicular to the sides of the triangle f c b, and 
make them equal to A h, and through their extremeties 
draw perpendiculars, cutting the circumference in the 
points i k, 1 m. Draw the lines i b, k f, If, me, and 
the horizontal projections of the three inferior faces 
obtained. The superior pyramid is similar and equal 
to the inferior, and solely opposed by its angles. De¬ 
scribe a circle passing through the three points of the 
first triangle, and draw within it a second equilateral 
triangle nop, of which the summits correspond to the 
middle of the faces of the former one. Each of these 
points will be the summit of a pentagon, as the points 
b c f. These pentagons have all their sides common, 
and it is only necessary therefore to determine one of 
these superior pentagons to have all the others. Six 
of the faces of the dodecahedron have now been pro¬ 
jected; the remaining six are obtained by joining the 
angular points already found, as q d, e r, t i, k s, &c. 

To obtain the vertical projeefion (No. Ill) begin 
with the three inferior faces. The point A in the 
horizontal projection being the summit of the inferior 
solid angle, will have its vertical in a; the points b c f, 
when raised to the height g G, will be in b c f, or simply 


54 


MODERN CARPENTRY 


b f. The points b gc being in a plane perpendicular to 
the vertical plane, will necessarily have the same ver¬ 
tical projection, b. The line a f will be the projection 
of the arris A f, and a b will be that of the arrises A b, 
A c, and of the line A g, or rather that of the triangle 
Abe, which is in a plane perpendicular to the vertical 
plane. But this triangle is only a protion of the given 
pentagonal face (No. 1), of which ATI is the perpen¬ 
dicular let fall from A on the side ED. Produce ab 
to e, making a e equal to A IT; e is the vertical projec¬ 
tion of the arris e d. This arris is common to the inferior 
pentagon, and to the superior pentagon e d q p r, which 
is also perpendicular to the vertical plane, and, conse¬ 
quently, its vertical projection will be e p, equal to a e. 

This projection can now be obtained by raising a 
vertical line through p, the summit of the superior 
pentagon, and from e as a centre, and with the radius 
A II or A H, describing an arc cutting this line in p, 
the point sought. But p n o belong to the base of the 
superior pyramid; therefore, if a perpendicular is 
drawn from n through y z to n, and the height p is trans¬ 
ferred to n by drawing through p' a line parallel to y z, 
n will be the projection of the points n and o. Through 
n draw s n a parallel to a e cutting perpendiculars 
drawn through s and A in the horizontal projection. 
Through s draw sf parallel to p e, and join a f, ap; 
set off on the perpendicular from r the height of s above 
y z at r, and draw r t parallel to y z, cutting the per¬ 
pendicular from t, and join n t. Draw perpendiculars 
from k and i though y z to k and i, make k and i the 
same height as e, and draw k i, and join i b, it. The 
vertical projection is now complete. 


SOLID GEOMETRY 


55 


XXII. In a given sphere to inscribe a tetrahedron, a 
hexahedron or cube, an octahedron and a dodecahed¬ 
ron. 

Let A B (Fig. 30) be the diameter of the given 
sphere. Divide it into three equal parts, D B being one 
of these parts. Draw D E perpendicular to A B, and 


H 



draw the chords A E, E B. A E is the arris of the tetra¬ 
hedron, and E B the arris of the hexahedron or cube. 
From the centre C draw the perpendicular radius C F, 
and the chord F B is the arris of the octahedron. Divide 
B E in extreme and mean proportion in G, and B G 
is the arris of the dodecahedron. The arrises being 
known, the solids can be drawn by the help of the prob¬ 
lems already solved. Draw the tangent A H equal to 
AB; join H C and AI; A I is the arris of an icosahe- 








56 


MODERN CARPENTRY 


dron which can be inscribed in the sphere, an icosa¬ 
hedron being a solid with twenty equal sides, all of 
which are equilateral triangles. See Fig. 30 y 2 . 



Fiff. 30% 













SOLID GEOMETRY 


57 


4. THE CYLINDER, CONE, AND SPHERE. 

XXIII. The horizontal projection of the cylinder, 
the axis of which is perpendicular to the horizontal 
plane, being' given to find the vertical projection. 

Let the circle ABCD (Fig. 31) be the base of the 
cylinder, and also its horizontal projection. From the 
points A and C raise perpendiculars to the gronnd-line 



B 

Fig. 31. 


a c, and produce them to the height of the cylinder— 
say, for example, a e, c f. Draw e f parallel to a c, and 
the rectangle a e f c is the vertical projection required. 

XXIV. Given' the traces of an oblique plane, to de¬ 
termine the inclination of the plane to both the H. P. 
and the V. P. 










58 


MODERN CARPENTRY 


Let v t and h t (Fig. 32) be the traces of the given 
plane. Draw the projections of a semi-cone having its 
axis a' b' in the vertical plane, the apex a' in the given 
v t and its base (a semi-circle) c e d in the h p and 
lying tangentially to the given h t. Then the base angle 



(0) of the cone gives the inclination of the plane to the 
h p. To determine the inclination of the plane to the 
v. p., draw the projections of a second semi-cone, hav¬ 
ing the axis m n in the h p, and the apex m in the 
given h t, while the base is in the v p and tangential 
to the vt. The base angle (0) of this cone gives the 
inclination to the v p. 

XXV. The base of a cylinder being given, and also 
the angles which the base makes with the planes cf 
projection, to construct the projections of the cylinder. 

Let the circle AGBII (Fig. 33) be the given base, 
and let each of the given angles be 45°. Draw the 
diamater AB making an angle of 45° with the ground- 



SOLID GEOMETRY 


59 


line or vertical plane, and draw the line A B, making 
with AB the given angle; and from A as a centre, 
with A B and A C as radii, describe arcs cutting A B in 
B and C. Then draw A D, B E perpendicular to A B, 



and equal to the length of the cylinder; the rectangle 
A E is the vertical projection of the cylinder parallel 
to the vertical plane and inclined to the horizontal 
plane A B in an angle of 45°. Now prolong indefinitely 
the diameter B A, and this line will represent the pro¬ 
jection on the horizontal plane of the line in which the 











60 


MODERN CARPENTRY 

generating circle moves to produce the cylinder. If 
from B and C perpendiculars be let fall on A B, k will 
be the horizontal projection of B, A k that of the 
diameter A B, and c that of the centre C. Through c 
draw h g perpendicular to A B, and make c h, c g equal 
to CH, G G; and the two diameters of the ellipse, 
which is the projection of the base of the cylinder, will 
be obtained, namely, A k and h g. 

In like manner, draw D F E the lines D d, F f, E c, 
perpendicular to the diameter A B produced, and their 
Intersections with the diameter and the sides of the 
cylinder will give the means of drawing the ellipse 
which forms the projection of the farther end of the 
cylinder. The ellipses may also be found by taking any 
number of points in the generating circle as I J, and 
obtaining their projections ij. The method of doing 
this, and also of drawing the vertical projection c f, 
will be understood without further explanation. 

XXVI. A point in one of the projections of a cone 
being given, to find it in the other projection. 

Let a (Fig. 34) be the given point. This point be¬ 
longs equally to the circle which is a section of the 
cone by a plane passing through the point parallel to 
the base, and to a straight line forming one of the 
sides of a triangle which is the section of the cone by a 
plane perpendicular to its base and passing through 
its vertex and through the given point, and of which 
fag is the horizontal, and fag the vertical projec¬ 
tion. To find the vertical projection of a, through a 
draw a a perpendicular to b c, and its intersection with 
f g is the point required; and reciprocally, a in the hori- 




SOLID GEOMETRY 61 

zontal projection may be found from a in the vertical 
projection, in the same manner. 

Otherwise, through a, in the horizontal projection, 
describe the circle a d c, and draw e e or c c, cutting 



the sides of the cone in e and c; draw c e parallel to 
the base, and draw a a, cutting it in a, the point re¬ 
quired. . 










62 MODERN CARPENTRY 

XXVII. On a given cylinder to describe a helix. 

Let abed, &c. (Fig. 35), be the horizontal projec¬ 
tion of the given cylinder. Take on this curve a series 
of equal distances, ab, be c d, &c., and through each 


n 



of the points a, b, c, &c. draw a vertical line, and pro-, 
duce it along the vertical projection of the cylinder. 
Then conceive a curve cutting all these verticals in the 
points a b c d. in such a manner that the height of the 
point above the ground-line may be in constant rela¬ 
tion to the arcs a b, b c, c d; for example, that a may be 



















SOLID GEOMETRY 


63 

the zero of height, that bb may be 1, c c 2, d d3, &c.; 
then this curve is named a helix. To construct this curve, 
carry on the vertical projection on each vertical line 
such a height as has been determined, as 1 on b, 2 on c, 
3 on d; and through these points will pass the curve 
sought. It is easy to see that the curve so traced is in¬ 
dependent of the cylinder on which it has been sup¬ 
posed to be traced; and that if it be isolated, its hori¬ 
zontal projection will be a circle. The helix is named 



after the curve which is its horizontal projection. Thus 
the helix in the example is a helix with a circular base. 
The vertical line f n is the axis of the helix, and the 



























64 


MODERN CARPENTRY 


height b b, comprised between two consecutive inter¬ 
sections of the curve with a vertical, is the pitch of the 
helix. 

XXVIII. On a given cone to describe a helix. 

Let the projections of the given cone be as shown in 
.(Fig. 36.) Divide the base of the cone in the horizon¬ 
tal projection into any number of equal parts, as a b, 
b e, c d, &c., and draw lines from the vertex to the 
points thus obtained. Set off along these lines a series 
of distances increasing in constant ratio, as 1 at b, 2 
at c, 3 at d, &c. The curve then drawn through these 
points, when supposed to be in the same plane, is 
called a spiral. If these points, in addition to approach¬ 
ing the centre in a constant ratio, are supposed also 
to rise above each other by a constant increase of 
height, a helical curve will be obtained on the vertical 
projection of the cone.* 

XXIX. A point in one of the projections of the 
sphere being given, to find it in the other projection. 

Let a be the given point in the horizontal projection 
of the sphere h b c i (Fig. 38). Any point on the sur¬ 
face of a sphere belongs to a circle of that sphere. 
Therefore, if a is the point, and a vertical plane b c is 
made to pass through that point to A B, the section of 


♦The octahedron is formed by the union of eight equilateral 
triangles; or, more correctly, by the union of two pyramids 
with square bases, opposed base to base, and of which all the 
solid angles touch a sphere in which they may be inscribed. 

It is essential that the diagram should be clearly seen as a 
solid, and not as a mere set of lines in one plane. Imagine h 
as the apex of one pyramid on the base kcBi, and a as the 
apex of the other pyramid on the opposite side of the same 
base. The octahedron is shown to be lying on its side abc. 



SOLID GEOMETRY 


65 


the sphere by this plane will be a circle, whose diam¬ 
eter will be b c, and the radius consequently, d b or d c; 
and the point a will necessarily be in the circumfer¬ 
ence of this circle. Since the centre d of this circle is 
situated on the horizontal axis of the sphere, and as 




this axis is perpendicular to the vertical plane, its ver¬ 
tical projection will be the point d. It is evident that 
the vertical projection of the given point a will be 
found in the circumference of the circle described from 
d with the radius d b or d c and at that point of it 
where it is intersected by the line drawn through a, per- 




























66 


MODERN CARPENTRY 


pendicular to A B. Its vertical projection will there¬ 
fore be either a or a, according as the point a is on the 
superior or inferior semi-surface of the sphere. 

The projection of the point may also be found thus: 
Conceive the sphere cut by a plane parallel to the hori¬ 
zontal plane of projection passing through the given 
point a. The resulting section will be the horizontal 
circle described from k, with the radius k a; and the 
vertical projection of this section will be the straight 
line g e, or g e; and the intersections of these lines with 
the perpendicular drawn through a, will be the projec¬ 
tion of a, as before. 


5. SECTIONS OF SOLIDS. 

To draw sections of any solid requires little more 
# 

than the application of the method described in the 
foregoing problems. Innumerable examples might be 
given, but a few selected ones will suffice. 

XXX. The projections of a regular tetrahedron be¬ 
ing given, to draw the section made by a plans perpen¬ 
dicular to the vertical plane and inclined to the hori¬ 
zontal plane. 

Let A B C D and abed (Fig. 39) be the given pro¬ 
jections, and E F G the given plane perpendicular to the 
vertical plane and inclined to the horizontal plane at 
an angle of 30°. The horizontal projection e f g of the 
section is easily found as shown. To find the correct 
section draw through e, f, and g, lines parallel to the 



SOLID GEOMETRY 


67 


ground-line A C, and a e one of them as E e set off the 
distances E F, E G at E f and E g, and through f and g 
draw perpendiculars cutting the other lines in F and 



G'. Join EG, GF, and F E. EFG is the correct sec¬ 
tion made by the plane. 

XXXI. The projections of a hexagonal pyramid be¬ 
ing given, to draw the section made by a plane perpen¬ 
dicular to the vertical plane and inclined to the hori¬ 
zontal plane. 

Let A B C D and a b c d e f g (Fig. 40) be the given 
projections, and III JK the given plane. The horizon- 














68 


MODERN CARPENTRY 


tal projection h i j k 1 m of the section is easily found 
as shown. Through m, 1, h, j, and i draw lines parallel 
to the ground-line A D, and on one of them as h m, set 
off the distances II I, II J, II K, at h M, h 1, li k. From 



h, M, 1, and k draw perpendiculars meeting the other 
lines in H, I, L, J, and K, and join the points of inter¬ 
section. III J K L M is the true section made by the 

plane. 























SOLID GEOMETRY 


69 


XXXII. The projections of an octagonal pyramid 
being given, to draw the section made by a vertical 
plane. 

Let A B C D F and a b c d e f (Fig. 41) be the given 
projections, and g h i j k of the section made by the 
plane is easily found by drawing g g, h h, i i, &c. per¬ 



pendicular to the ground-line AD. On A D produced 
set off the distances g h, h i, i j, and j k at G h, h i, &c., 
and from the points thus found draw perpendiculars 
to G Iv meeting lines drawn from h, i, and j parallel to 
A D, in IT, I, and J. Join G H, H I, I J, and J K. G HI 
J K is the true section made by the plane. 

























70 


MODERN CARPENTRY 


A cylinder may be cut by a plane in three different 
ways—1st, the plane may be parallel to the axis; 2nd, it 
may be parallel to the base; 3rd, it may be oblique to 
the axis or the base. 

In the first case, the section is a parallelogram, whose 
length will be equal to the length of the cylinder, and 
whose width will be equal to the chord of the circle of 
the base in the line of section. Whence it follows, that 
the largest section of this kind will be that made by a 
plane passing through the axis; and the smallest will 
when the section plane is a tangent—the section in that 
case will be a straight line. 

When the section plane is parallel to the base, the 
section will be a circle equal to the base. When the 
section plane is oblique to the axis or the base, the 
section will be an ellipse. 



XXXIII. To draw the section of a cylinder by a 
plane oblique to the axis. 

Let ABCD (Fig. 42) be the projection of a cylin¬ 
der, of which the circle EH FK represents the base 
divided into twenty equal parts at b c d e, &c., and let it 
be required to draw the section made by the plane A C. 



























SOLID GEOMETRY 


71 


The circular base must be drawn in such a position that 
the axis of the cylinder when produced meets the centre 
of the circle. Through the centre of the circle draw 
the diameter II K perpendicular to the axis produced. 
Then through the divisions of the base, bed, &c. draw 
lines parallel to the axis, and meeting the section plane 
in 1, 2, 3, &c., and through these points draw perpen¬ 
diculars to A C making them equal to the correspond¬ 
ing perpendiculars from II K, i e, lb, 2 c, 3 d, &c. A 
curve drawn through the points thus found will be an 
ellipse, the true section of A B C D on the plane A C.* 

The Cone. A cone may be cut by a plane in five dif¬ 
ferent ways, producing what are called the conic sec¬ 
tions: 1st. If it is cut by a plane passing through the 
axis, the section is a triangle, having the axis of a 
cone as its height, the diameter of the base for its base, 
and the sides for its sides. If the plane passes through 
the vertex, without passing through the axis, as c e 
(Fig. 43), the section will still be a triangle, having 
for its base the chord c e, for its altitude the line c e, and 
for its sides the sides of the cone, of which the lines 
c e, o e are the horizontal and the line c e the vertical 
projections. 2nd. If the cone is cut parallel to the base, 
as in g h, the section will be circle, of which g h will be 
the diameter. 3rd. When the section plane is oblique 
to the axis, and passes through the opposite sides of 
the cone, as m p h the section will be an ellipse, m n li. 
4th. When the plane is parallel to one of the sides of 
the cone, as r h, the resulting section is a parabola r s 
h t u. 5th. When the section plane is such as to pass 
through the sides of another cone formed by producing 

*The point is assumed to be in the inferior half of the 
cylinder. 



72 


MODERN CARPENTRY 


the sides of the first beyond the vertex, as the plane 
q h, the resulting curve in each cone is a hyperbola. 



Several methods of drawing the curves of the conic 
sections have already been given in Plane Geometry, 




























SOLID GEOMETRY 


73 


Vol. 1. Hepe their projections, as resulting from the 
sections of the solid by planes, are to be considered. 
If the mode of finding the projections of a point on the 
surface of a given cone be understood, the projections 
of the curves of the conic sections will offer no 
difficulty. Let the problem be: First, to find the pro¬ 
jections of the sectioij made by the plane mh. Take 
at pleasure upon the plane the several points, as p, &c. 
Let fall from these points perpendiculars to the hori¬ 
zontal plane, and on these will be found the horizontal 
projection of the points; thus, in regard to the point p 

-. Draw through p a line parallel to A B: this line 

will be the vertical projection of the horizontal plane 
cutting the cone, and its horizontal projection will be 
a circle, with s n for its radius. With this radius, 
therefore, from the centre e, describe a circle cutting, 
twice, the perpendicular let fall from p; the points of 
intersection will be two points in the horizontal projec¬ 
tion of the circumference of the ellipse. In the same 
manner, any other points may be obtained in its cir¬ 
cumference. The operation may often be abridged by 
taking the point p in the middle of the line m h; for 
then m h will be the horizontal projection of the major 
axis, and the two points found on the perpendicular 
let fall from the central point p will give the minor 
axis. 

To obtain the projections of the parabola, more 
points are required, such as r, 2, s, 3, h, but the mode 
of procedure is the same as for the ellipse. The ver¬ 
tical projection of the parabola u s h t u is shown at 

u s h t u. 



74 


MODERN CARPENTRY 


The projections of the section plane which produces 
the hyperbola are in this case straight lines, q h, z li. 

XXXIV. To draw the section of a cone made by a 
plane cutting both its sides, i. e., an ellipse. 

Let A D B (Fig. 44) be the vertical projection of 
the cone A C B the horizontal projection of half its 

D 



base, and EF the line of section.' From the points E 
and F let fall on A B the perpendiculars E G, FII. 
Take any points in E F, as k, 1, m, n, &c., and from 
them draw lines parallel to A B, as k p, 1 q, m r, &c.. 































SOLID GEOMETRY 


7S 


and also lines perpendicular to AB, as li 1, 12 m3, &c. 
Also from p, q, r, &c., let fall perpendiculars on A B, 
namely, p a, q z, r y, &c. From the centre of the base 
of the cone, 1, with radius la, lz, ly, &c., describe arcs 
cutting the perpendiculars let fall from k, 1, m, &c., in 

l, 2, 3, &c. A curve traced through these points will 
be the horizontal projection of the section made by the 
plane E F. To find the true section,—Through k, 1, 

m, &c., draw kt, lu, m v, n w, perpendicular through 
E F, and make them respectively equal to the corres¬ 
ponding ordinates, 5 1, 6 2, 7 3, &c., of the horizontal 
projection G4H, and points will be obtained through 
which the half E w F of the required ellipse can be 
traced. It is obvious that, practically it is necessary 
only to find the minor axis of the ellipse, the major 
axis E F being given. 

XXXV. To draw the section of a cone made by a 
plane parallel to one of its sides, i. e. a parabola. 

Let A D B (Fig. 45) be the vertical projection of a 
right cone, and A C B half the plan of its base; and lei 
E F be the line of section. In E F take any number of 
points, E. a, b, c, e, F, and through them draw lines 
E IT, a 61, b 7 2, &c., perpendicular to A B, and also 
lines parallel to A B, meeting the side of the cone in 
f, g, h, k, 1: from these let fall perpendiculars on A B, 
meeting it in m n, o, p, q. From the centre of the base 
1, with the radii lm, In, lo, &c., describe arcs cutting 
the perpendiculars let fall from the section line in the 
points 1, 2, 3, 4, 5; and through the points of intersec¬ 
tion trace the line II 12345G, which is the horizontal 
projection of the section. To find the true section, 
from E, a, b, c, d,e, raise perpendiculars to E F, and 


76 


MODERN CARPENTRY 


make them respectively equal to the ordinates in the 
horizontal projection, as E r equal to EII, as equal to 
6 1, &c., and the points rstuvw in the curve will be 


D 



Fig. 45. 


obtained. The other half of the parabola can be drawn 
by producing the ordinates we, v d, &c., and setting 
the same distances to the right of E P. 

XXXVI. To draw the section of a cone made by a 
plane parallel to the axis i. e. an hyperbola. 























SOLID GEOMETRY 


77 


Let cl c d (Fig. 46) be the vertical projection of the 
cone, cl cj r d one half of the horizontal projection of 
the base, and q r the section plane. Divide the line r q 
into any number of equal parts in 1, 2, 3, h, &c., and 
through them draw lines perpendicular to d cl. From 
c as centre, with the radii cl, c2, &c., describe the arcs 

% 


f 

c 



cutting dd; and from the points of intersection draw 
perpendiculars cutting the sides of the cone in 1, 2, 3, 
and these heights transferred to the corresponding per¬ 
pendiculars drawn directly from the points 1, 2, 3, &c., 
in r q, will give points in the curve. 

















78 


MODERN CARPENTRY 


XXXVII. To draw the section of a cuneoid made 
by a plane cutting both its sides. 

Let ACB (No. 1, Fig. 47) be the vertical projection 
of the cuneoid, and A 5 B the plan of its base, and 
A B (No. 4) the length of the arris at C, and let D E 
be the line of section. Divide the semicircle of the 
base into any number of parts 1, 2, 3, 4, 5, &c., and 
through them draw perpendiculars to A B, cutting it 
in 1, m, n, o, p, &c., and join Cl, C m, Cn, &c., by lines 
cutting the section line in 6, 7, 8, 9, &c. From these 
pointy draw lines perpendicular to D E, and make 
them equal to the corresponding ordinates of the semi¬ 
circle, either by transferring the lengths by the com¬ 
passes, or by proceeding as shown in the figure. The 
curve drawn through the points thus obtained will give 
the required section. 

The section on the line D K is shown in No. 2, in 
which A B equals D K; and the divisions e f g h k in 
D K, &c., are transferred to the corresponding points 
on AB; and the ordinates el, f m, g n, &c., are made 
equal to the corresponding ordinates 11, m2, n3, of the 
semicircle of the base. In like manner, the section of 
the line G II, shown in No. 3, is drawn.* 

XXXVIII. To describe the section of a cylinder 
made by a curve cutting the cylinder. 

Let A BD E (Fig. 48) be the projection of the cylin¬ 
der, and C D the line of the section required. On AB 


*A cuneoid is a solid ending in a straight line, in which, if 
any point be taken, a perpendicular from that point may’be 
made to coincide with the surface. The base of the cuneoid 
may be of any form; but in architecture it is usually semi¬ 
circular or semi-elliptical, and parallel to the straight line 
forming the other end. 



SOLID GEOMETRY 


79 


describe a semicircle, and divide it into any number of 
parts. From the points of division draw ordinates 1 h, 
2k, 31, 4m, &c., and produce them to meet the line of 
section in o, p, q, r, s, t, u, v, w. Bend a rule or slip 
of paper to the line C D, and prick off on its points C, 
o, p, q, &c.; then draw any straight line F G, and, un- 



Fig\ 47. 

bending the rule, transfer the points C, o, p, q, &c., to 
F, a, b. c, d, &c. Draw the ordinates al, b 2, c 3, &c., 
and make them respectively equal to the ordinates h 1, 
k2, ] 3, &c., and through the points found trace the 


curve. 









































80 


MODERN CARPENTRY 


XXXIX, To describe the section of a sphere. 

Let A B D C (Fig. 49) be the great circle of a sphere, 
and F G the line of the section required. Then since all 
the sections of a globe or sphere are circles, on F G de¬ 
scribe a semicircle F 4G, which will be the section re¬ 
quired. 



Or, in F G take any number of points, as m, 1, k, H, 
and from the centre of the great circle E, describe the 
arcs H n, k o, 1 p, m q, and draw the ordinates H 4, k 3, 
1 2, ml, and n 4, o3, p2, q 1; then make the ordinates on 
F G equal to those on B C, and the points so obtained 
will give the section required.* 

*The projections of sections of spheres are, if the section 
panes are oblique, either straight lines or ellipses, and are 
found as follows: 

Let ab (Fig. 50) be the horizontal projection of the sec¬ 
tion plane. On the line of section take any number of 
points, as a, c, b. and through each of them draw a line per¬ 
pendicular to y k. 



























SOLID GEOMETRY 


81 


XL. To describe the section of an ellipsoid when a 
section through the fixed axis, and the position of the 
line of the required section are given. 

Let ABCD (Fig. 51) be the section through the 
fixed axis of the ellipsoid, and F G the position of the 
line of the required section. Through the centre of 



the ellipsoid draw BD parallel to FG; bisect FG in 
H and draw A C perpendicular to FG; join B C, and 
from F draw F K, parallel to B C, and cutting A C pro¬ 
duced in K; and then will HK be the height of the 
semi-ellipse forming the section on F G. 























82 


MODERN CARPENTRY 


Or, the section may be found by the method of or¬ 
dinates, thus: As the section of the ellipsoid on the 
line A C is a circle, from the point of intersection of 
B D and A C describe a semicircle A E C. Then on 
H G, the line of section, take any number of points, 1, 
m, p, and from them raise perpendiculars cutting the 
ellipse in q, r, s. From q, r, s draw lines perpendicular 
to A C, cutting it in the points 4, 5, 6; and again, from 
the intersection of B D and A C as centre draw the 
arcs 41, 5 m, 6n, Co, cutting IT G in 1, m, n, o; then 
II o, set off on the perpendicular from H to K, is the 
height of the section; and the heights H n, IT m, HI, set 
off on the perpendiculars from ' to 3, n to 2 and p to 1, 
give the heights of the ordinates. 

XLI. To find the section of a cylindrical ring per¬ 
pendicular to the plane passing through the axis of the 
ring, the line of section being given. 

Let ABED (Fig. 52) be the section through the 
axis of the ring, AB a straight line passing through 
the concentric circles to the centre C, and D E be the 
line of section. On A B describe a semicircle; take in 
its circumference any points as 1, 2, 3, 4, 5, &c., and 
draw the ordinates 1 f, 2g, 3 h, 4 k, &c. Through the 
points f, g, h, k, 1, &c., where the ordinates meet the 
line A B, and from the centre C, draw concentric 
circles, cutting the section line in m, n, o, p, q, &c. 
Through these points draw the lines ml, n2 / o 3, &c., 
perpendicular to the section line, and transfer to them 
the heights of the ordinates of the semicircle f 1, g 2, 
&c.; then through the points 1, 2, 3, 4, &c., draw the 
curve D 5 E, which is the section required. 


SOLID GEOMETRY 


83 


Again, let R S be the line of the required section; 
then from the points t, n, v, w, c, x, d, &c., where the 
concentric circles cut this line, draw the lines 11, u2, 
v 3, &c., perpendicular to R S, and transfer to them the 



corresponding ordinates of the semicircle; and through 
the points 1, 2 , 3 , 4, e, 5, f, &c., draw the curve R e f S, 
which is the section required. 

XLII. To describe the section of a solid of resolu¬ 
tion the generating curve of which is an agee. 

Let A D B (Fig. 53) be half the plan or base of the 
solid, A a b B the vertical section through its axis, and 
E F the line of section required. From G draw C 5 
perpendicular to E F, and bisecting it in m. In E m 


























84 


MODERN CARPENTRY 


take any number of points, g h k, &c., and through 
them draw the lines g 1, li 2, k 3, &c., perpendicular to 
E F. Then from C as a centre, through the points g, 
h, k, &c., draw concentric arcs cutting A B in r, s, t, u, 
v, and through these points draw the ordinates r 5, s 4, 
t 3, &c., perpendicular to A B. Transfer the heights 
of the ordinates on A B to the corresponding ordinates 
on each side of the centre of EF; and through the 
points 1, 2, 3, 4, 5, &c., draw the curve E 5 F which is 
the section required. 


E 



XLIII. To find the section of a solid of resolution, 
the generating curve of which is of a lancet form. 

Let A D B (Fig. 54) be the plan of half the base, A 
E B the vertical section, and F G the line of the re¬ 
quired section. The manner of finding the ordinates 


















SOLID GEOMETRY 


85 


and transferring the heights is precisely the same as in 
the last problem. 

XLIV. To find the section of an ogee pyramid with 
a hexagonal base. 

Let ADEPB (Fig. 55) be the plan of the base oJ: 
the pyramid, A a b B a vertical section through its 
axis, and G H the line of the required section. Draw 



the arrises CD, C E, C F. On the line of section G H, 
at the points of intersection of the arrises with it, and 
at some intermediate points k, m, o, q (the correspond¬ 
ing points k and q, and m and o, being equidistant 
from n), raise indefinite perpendiculars. Through 
these points k, 1, m, n, o, p, q, draw lines parallel to 
the sides of the base, as shown by dotted lines; and 
from the points where these parallels meet the line 
A B, draw r 4, s 3, t 2, u 1, perpendicular to A B. These 
perpendiculars transferred to the ordinates n 4, m3, 
o 5, 1 2, p 6, k 1, q 7, will give the points 1, 2, 3, 4, 4, 5, 
6, 7, through which the section can be drawn. 
















86 


MODERN CARPENTRY 


6. INTERSECTIONS OF CURVED SURFACES. 

When two solids having curved surfaces penetrate 
or intersect each other, the intersections of their sur¬ 
faces form curved lines of various kinds. Some of 
these, as the circle, the ellipse, &c., can be obtained in 
the plane; but the others cannot, and are named curves 
of double curvature. The solution of the following 
problems depends chiefly on the knowledge of how to 
obtain, in the most advantageous manner, the pro- 
pections of a point on a curved surface; and is in fact 
the application of the principles elucidated in the sev¬ 
eral previous problems. The manner of constructing the 
intersections of these curved surfaces which is the 
simplest and most general in its application, consists in 
conceiving the solids to which they belong as cut by 
planes according to certain conditions, more or less 
dependent on the nature of the surfaces. These sec¬ 
tion planes may be drawn parallel to one of the planes 
of the projection; and as all the points of intersection 
of the surfaces are found in the section planes, or on 
one of their projections, it is always easy to construct 
the curves by transferring these points to the other 
projection of the planes. 

XLV. The projection of two equal cylinders which 
intersect at right angles being given, to find the pro¬ 
jections of their intersections. 

Conceive, in the horizontal projection (Fig. 56), a 
series of vertical planes cutting the cylinders parallel 
to their axis. The vertical projections of all the sec¬ 
tions will be so many right-angled parallelograms, sim- 


SOLID GEOMETRY 


87 




ilar to e f e f, which is the result of the section of the 
cylinder from surface to surface. The circumference 
of the second cylinder, whose axis is vertical, is cut by 
the same plane, which meets its upper surface at the 
two points g, h, and its under surface at two corres¬ 
ponding points. The vertical projections of these 
points are on the lines perpendicular to a b, raised on 



Fig. 56. 


each of them, so that upon the lines e f, e f, will be sit¬ 
uated the intersections of these lines at the points g, 
h, and g, h and the same with other points i, k, 1, m. It 
is not necessary to draw a plan to find these projec¬ 
tions. All that is actually required is to draw the 
circle representing one of the bases (as no) of the 
cylinder laid flat on the horizontal plane. Then to 
produce gh till it cuts the circle at the superior and 



































88 


MODERN CARPENTRY 


inferior points G, G, and to take the heights eG, e G, 
and carry them, upon a . b, from g to g g and from h to 
h, h. 

Fig. 57 is the vertical projection made on the line 
x z. 



XLVI. To construct the projections of two unequal 
cylinders whose axis intersect each other obliquely. 

Let A (Fig. 58) be the vertical projection of the two 
cylinders, and h S d e the horizontal projection of their 
axis Conceive in the vertical projection, the cylin¬ 
ders cut by any number of horizontal planes; the hori¬ 
zontal projections of these planes will be rectangles, 
as in the previous example, and their sides will be 
parallel to the axis of the cylinders. 

The points of intersection of these lines will be the 
points sought. Without any previous operation, six of 
those points of intersection can be obtained. For ex¬ 
ample the point c is situated on d e, the highest point 
of the smaller cylinder; consequently, the horizontal 
projection of c is on d e, the horizontal projection of 




















































SOLID GEOMETRY 


89 


cl e, and it is also on the perpendicular let fall from c, 
that is to say, on the line c f parallel to the axis of the 
cylinder S h. The point sought will therefore, be the 
intersection of those lines at c. In the same way i is 



obtained. The point j is on the line k 1, which is in the 
horizontal plane passing through the axis d e, the hori¬ 
zontal projections of kl are k 1, and its opposite mn; 
therefore in letting fall perpendiculars from j p, the 
intersections of these with k 1, m n, give the points j j, 






















90 


MODERN CARPENTRY 


p p. Thus six points are obtained. Take at pleasure 
an intermediate point q; through this point draw a 
line rs parallel to ah, which will be the vertical pro¬ 
jection of a horizontal plane cutting the cylinder in 
q. The horizontal projection of this section will be, as 
in the preceding examples, a rectangle which is ob¬ 
tained by taking, in the vertical projection, the height 
of the section plane above the axis d e, and carrying it 
on the base in the horizontal projection from G to T. 
Through T is then to be drawn the line Q U perpendic¬ 
ular to G T; and through Q and U the lines parallel to 
the axis; and the points in which these lines are inter¬ 
sected by the perpendiculars let fall from q u are the 
intermediate point required. Any number of interme¬ 
diate points can thus be obtained; and the curve being 
drawn through them, the operation is completed. 

XLVII. To find the intersections of a sphere and a 
cylinder. 

Let e f c d and i k g h (Fig. 59) be the horizontal pro¬ 
jections of the sphere and cylinder respectively. Draw 
parallel to A B, as many vertical section planes are 
considered necessary, as e f, c d. These planes cut at 
the same time both the sphere and the cylinder, and the 
result of each section will be a circle in the case of the 
sphere, and a rectangle in the case of the cylinder. 
Through each of the points of intersection g, h, i, k, and 
from the centre 1, draw indefinite lines perpendicular 
to A B. Take the radius of the circles of the sphere 
proper to each of these sections and with them, from 
the centre 1, cut the correspondent perpendiculars in 
g g, h h, i i, &c., and draw through these points the 
curves of intersection. 


SOLID GEOMETRY 91 

XLVIII. To construct the intersection of two right 
cones with circular bases. 

The solution of this problem is founded on the 
knowledge of the means of obtaining on one of the 
projections of a cone a point given on the other. 



No. 1. Let A B (Fig. 60) be the common section of 
the two planes of projection, the circles g cl e f and g h 
ik the horizontal projections of the given cones, and 
the triangles dif and hlk their vertical projections. 
Suppose these cones cut by a series of horizontal planes: 






















92 MODERN CARPENTRY 

each section will consist of two circles, the intersections 
of which will be points of intersection of the conical 
surfaces. For example, the section made by a plane 
m n will have for its horizontal projections two circles 
of different diameters, the radius of the one being i m, 
and of the other 1 o. The intersecting points of these 
are p and q, and these points are common to the two 
circumferences: and their vertical projection on the 
plane mn will be p q. Thus, as many points may be 
found as is necessary to complete the curve. 

But there are certain points of intersection which 
cannot be rigorously established by this method with¬ 
out a great deal of manipulation. The point r in the 
figure is one of those; for it will be seen that at that 
point the two circles must be tangents to each other, 
and it would be difficult to fix the place of the 
section plane s t so exactly by trial, that it would just 
pass through the point. 

It will be seen that the point r must be situated in 
the horizontal projection of the line g i a perpendicu¬ 
lar i 1 equal to the height of the cone. From one raise 
a perpendicular and make it equal to the height of the 
second cone, and draw its side Li; and from the point 
of intersection R let fall a perpendicular on g i meet¬ 
ing it in r; through r draw an indefinite line perpendic¬ 
ular to A B, and set up on it from A B to r the height 
rR. The point r can also be obtained directly in the 
vertical projection by joining i g and 1 i as shown. 

Bisect rg in u, and from u as a centre, with the 
radius u r, describe a circle, the circumference of 
which will be the horizontal projection of the intersec¬ 
tion of the two cones. The vertical projection of this 


SOLID GEOMETRY 


93 


circle will be g p r q, and can be found by the method 
indicated above. 

No. 2. Conceive the horizontal projection (Fig. 61, 
No. 1) a vertical plane C D cutting both cones through 
their axis: the sections will be two triangles, having 
the diameters of the bases of the cones as their bases, 



Fig. 61. 


and the height of the cones as their height. And as in 
the example the cones are equal, the triangles will also 
be equal, as the triangles c e f, g f d, in the vertical pro¬ 
jection. Conceive now a number of inclined planes, as 
c n m, c k p, &c., passing through the different points 
of the base, but still passing through the summits of 
the cones: the sections which result will still be tri¬ 
angles (as has already been demonstrated), whose 






























94 


MODERN CARPENTRY 


bases diminish in proportion as the planes recede from 
the centres of the bases of the cones, until at length 
the plane becomes a tangent to both cones and the re¬ 
sult is a tangent line whose projections are h c, li c, g e, 
f f. It will be observed that the circumferences of the 
bases cut each other at m and i, which are the first 
points of their intersections, whose vertical projections 
are the point m merely. If the projections of the other 
points of intersection on the lines of the section planes 
are found (an operation presenting no difficulty, and 
easily understood by the inspection of the figure), it 
will be seen that the triangles n c m, mco, k e p, q e r, 
&c., in the horizontal projection, have for their vertical 
projections the triangles n e m, m o, k e p, &c., and that 
the intersections of the cones are in a plane perpendic¬ 
ular to both planes of projection, and the projections of 
the intersections are the right lines i m, m 3. From 
the known properties of the conic sections, the curve 
produced by this plane will be a hyperbola. Fig. 61, 
No. 2, gives the projections of the cones on the line o x. 

No. 3. The next example (Fig. 62) differs from 
the last in the inequality of the size of the cones. Sup¬ 
pose an indefinite line C D to be the horizontal projec¬ 
tion of the vertical section plane, cutting the two cones 
through their axis e f. Conceive in this plane an in¬ 
definite line e f D, passing through the summits of the 
cones, the vertical projection of this line will be efd: 
from d let fall on C D a perpendicular meeting it in D; 
this will be the point in which the line passing through 
the summits of the cones will meet the horizontal plane; 
and it is through this point, and through the summits 
e and f, that the inclined section planes should be made 


SOLID GEOMETRY 95 

to pass. The horizontal traces of these planes are 0 D, 
G D, &c.: 0 D is then the trace of a tangent plane to 
the two conical surfaces 0 e, P f; and the plane e G D 
cuts the greater cone, and forms by the section the tri¬ 
angles G e H in- the horizontal, and geh in the ver¬ 
tical projection; and it cuts the lesser cone, and forms 
the triangles 1 f J ; i f j. In the horizontal projection it 



is seen that the sides H e, 1 f of the triangles intersect 
in k, which is therefore the horizontal projection of one 
of the points of intersection; and its vertical projection 
is k. In the same manner, other points can be found. 
It is seen at once that M, N, 1, are also points in the in¬ 
tersection. The curves traced through the points M k 
] N in the horizontal, and m k 1 in the vertical projec¬ 
tion, are the projections of the intersection of the two 


Fig. 6L. 


cones. 







96 


MODERN CARPENTRY 


7. COVERINGS OF SOLIDS, 

I. Regular Polyhedrons. A solid angle cannot be 
formed with fewer than three plane angles. The sim¬ 
plest solid is therefore the tetrahedron or pyramid 
having an equilateral triangle for its base, and its other 
three sides formed of similar triangles. 

The development of this figure (Pig. 63) is made by 
drawing the triangler base ABC, and then drawing 
around it the triangles forming the inclined sides. If 
the diagram is on flexible material, such as paper, then 
cut out, and the triangles folded on the lines A B, B C, 
CA, the solid figure will be constructed. 



Fig. 63. Fig. 64. 


The hexahedron, or cube is composed of six equal 
squares (Fig. 64) ; the octahedron (Fig. 65) of eight 
equilateral triangles; the dodecahedron (Fig. 66) of 
twelve pentagons; the icosahedron (Fig. 67) of twenty 


















SOLID GEOMETRY 


97 


equilateral triangles. In these figures, A is the eleva¬ 
tion, and B the development. 

The elements of these solids are the equilateral tri¬ 
angle, the square, and the pentagon. The irregular 
polyhedrons may be formed from those named, by cut¬ 
ting off the solid angles. Thus, in cutting off the 



angles of the tetrahedron, there results a polyhedron 
of eight faces, composed of four hexagons and four 
equilateral triangles. The cutting off the angles 
of the cube, in the same manner gives polyhedron of 























98 


MODERN CARPENTRY 


fourteen faces, composed of six octagons, united by 
eight equilateral triangles. 

The same operation performed on the octahedron 
produces fourteen faces, of which eight are hexagonal 
and six square; on the dodecahedron it gives thirty-two 
sides, namely, twelve decagons, and twenty triangles; 
on the icosahedron it gives thirty-two sides—twelve 
tentagons and twenty hexagons. This last approaches 
almost to the globular form and can be rolled like a 
ball. 



Pig. 67. 


The other solids which have plane surfaces are the 
pyramids and prisms. These may be regular and irreg¬ 
ular: they may have their axis perpendicular or in¬ 
clined: they may be truncated or cut with a section, 
parallel or oblique, to their base. 






SOLID GEOMETRY 


99 


II. Pyramids. The development of a right pyra¬ 
mid, of which the base and the height are given, offers 
no difficulty. The operation consists (Fig. 68) in ele¬ 
vating on each side of the base, a triangle having its 
height equal to the inclined height of each side, or, 
otherwise, connecting the sides, together as shown by 
the dotted lines. 




In an oblique pyramid the development is found as 
follows: Let abed (Fig. 69) be the plan of the base 
of the pyramid, a b c d its horizontal projection, and 
E F G its vertical projection. Then on the side d c 
construct the triangle c R cl, making its height equal 
to the sloping side of the pyramid F G. This triangle 
is the development of the side cl P c of the pyramid. 

L. OF C. 

















100 


MODERN CARPENTRY 


Then from d, with the radius EE, describe an arc 0; 
and from R, with the radius equal to the true length 
e G oi the arris E G, describe another arc intersecting 
the last at 0. Join R 0, dO; the triangle dRO will 
be the development of the side a P d. In the same way, 
describe the triangle cRT, for the development of 
the side b P c. From R, again, with the same radius 
R 0 , describe an arc S, which intersect by an arc de¬ 
scribed from 0 with the radius a b; and the triangle 
0 R S will be the development of the side a P b. 

If the pyramid is truncated by a plane w a parallel 
to the base, the development of that line is obtained by 
setting off from R on R c, and R d, the true length of 
the arris G a in x and 2, and on R S, R O, and R T, the 
true length of the arris G w in 4, 3, 1; and drawing the 
lines 1 x, x2, 2 3, 3 4, parallel to the base of the respect¬ 
ive triangles T R c, c R d, dRO, 0 R S. If it is trunc¬ 
ated by a plane wy, perpendicular to the axis, then 
from the point R, with the radius equal to the true 
length of the arris G w, or G y, describe an arc 1 4, and 
inscribe in it the sides of the polygon forming the pyra¬ 
mid. 

III. Prisms. In a right prism, the faces being all 
perpendicular to the bases which truncate the solid, it 
results that their development is a rectangle composed 
of all the faces joined together, and bounded by two 
parallel lines equal in length to the contour of the 
bases. Thus, in Fig. 70, a b c d is the base, and b e the 
height of the prism; the four sides will form the rect¬ 
angle b e f g, and e h i k will be the top of the prism. 

4he full iines show another method of development. 


SOLID GEOMETRY 101 

When a prism is inclined, the faces form different 
angles with the lines of the contours of the bases; 
whence there results a development, the extremities 
of which are bounded by lines forming parts of poly¬ 
gons. 



K 



After having drawn the line C C (Fig. 71), which in¬ 
dicates the axis of the prism and the lines A B, D E, 
the surfaces which terminate it, describe on the middle 
of the axis the polygon forming the plan of the prism, 
taken perpendicularly to the axis, and indicated by the 
figures 1 and 8. Produce the sides 12, 6 5, parallel to 
the axis, until they meet the lines A B, D E. These 
lines then indicate the four arrises of the prism, cor- 
















102 MODERN CARPENTRY 

i 

responding to the angles 1 2 5 6. Through the points 
S 3 7 4 draw lines parallel to the axis meeting A B, D E 
in F H, G L. These lines represent the four arrises 
8 3 7 4. * 

In this profile the sides of the plan of the polygon 
12345678 give the width of the faces of the prism, 


2 S 



Fig. 71. 


and the lines AD, F II, G L, BE their length. From 
this profile can be drawn the horizontal projection, in 
the manner shown below. To trace the development of 
tins prism on a sheet of paper, so that it can be folded 

* In, Fig* .75 another example is given, but as the method of 
piocedure is the same as in Fig 71, detailed description is un¬ 
necessary. 





















SOLID GEOMETRY 


103 


together to form the solid, proceed thus: On the 
middle of C C raise an indefinite perpendicular M N. 
On that line set off the width of the faces of the prism, 
indicated by the polygon, in the points 01234567 8. 
Through these points draw lines parallel to the axis, 
and upon them set off the lengths of the lines in profile, 
thus: From the points 0, 1, and 8, set of the length 
M D in the points DDD; from 2 and 7, set off a II in 
H from 3 and 6, set off b L in L and L; and so on. 
Draw the lines DD, D H L E, EE, E L H D, for the 
contour of the upper part of the prism. To obtain the 
contour of the lower portion, set off the length M A 
from 0, 1, and 8 to AAA, the length aF from 2 and 
7 to F and F, the length b G from 3 and 6 to G and G, 
and so on; and draw A A, A F G B, B B, B G F A, to 
complete the contour. The development is completed 
by making on B B and E E the polygons 1 2 3 4 5 6 BB, 
1 2 3 4 5 6 EE, similar to the polygons of the horizontal 
projection. 

IV. Cylinders. Cylinders may be considered as 
prisms, of which the base is composed of an infinite 
number of sides. Thus we shall obtain graphically the 
development of a right cylinder by a rectangle of the 
same height, and of a length equal to the circumference 
of the circle, which serves as its base. 

To find the covering of a right cylinder. 

Let A B C D (Fig. 72) be the seat or generating sec¬ 
tion. On A D describe the semicircle A 5 D, represent¬ 
ing the vertical section of half the cylinder, and divide 
its circumference into any number of parts, 1, 2, 3, 4, 


104 


MODERN CARPENTRY 


5, &c., and transfer those divisions to the lines A D 
and B C produced; then the parallelogram DCGF will 
he the covering of one half the cylinder. 



To find the edge of the covering when it is oblique 
in regard to the sides of the cylinder. 

Let A B C D (Fig. 73) be the seat of the generating 
section the edge B C being oblique to the sides A B, 
D 0. Draw the semicircle A 5 D, and divide it into any 



number of parts as before; and through the divisions 
draw lines at right angles to A D, producing them to 
meet BCinrs, t, u, v, &c. produce A D, and the lines 
la, 2 b, 3 c, &c., perpendicular to D F. To these lines 

























































SOLID GEOMETRY 


105 


transfer the length of the corresponding lines inter¬ 
cepted between A D and B C, that is, to la transfer the 
length p z, to 2 b transfer o y, and so on, by drawing 
the lines z a, y b, x c, &c., parallel to A F. Through the 
points thus obtained, draw the curved line C a b c, &c., 
to G; then shall D F C G be the development of the 
covering of the semi-cylinder A B C D. 



Fig. 74. 


To find the covering of a cylinder contained between 
two oblique parallel planes. 

Let A B C D (Fig. 74) be the seat of the generating 
section. From A draw A G perpendicular to A B, and 
produce C D to meet it in E. On A E describe the semi¬ 
circle, and transfer its perimeter to E G, by dividing 
it into equal parts, and setting off corresponding di¬ 
visions on E G. Through the divisions of the semicircle 
draw lines at right angles to A E, producing them to 


































106 


MODERN CARPENTRY 


meet the lines A D and B C, in i, k, 1, m, &c. Through 
the divisions on E G draw lines perpendicular to it; 
then through the intersections of the ordinates of the 
semicircle, with the line A D, draw the lines i a, k z, 1 y, 


bdfk A 



Fig-. 75. 


&c., parallel to A G and where these intersect the per¬ 
pendiculars from E G, in the points a, z, y, x, w, u, &c., 
trace a curved line G D, and draw parallel to it the 
curved line 11 C; then will D C II G be the development 
of the covering of the semi-cylinder A B C D. 

To find the covering of a semi-cylindric surface 
bounded by two curved lines. 

The construction to obtain the developments of these 
coverings (Figs. 76 and 77) is precisely similar to that 
described in Fig. 74. 






SOLID GEOMETRY 


107 






Fig. 77, 




























































































108 MODERN CARPENTRY 

V. Cones. We have considered cylinders as prisms 
with polygonal bases. In a similar manner we may re¬ 
gard cones as pyramids. 

In right pyramids, with regular symmetrical bases, 

G' 


G 

Fig. 7S. 

as the lines of the arrises extending from the summit 
to the base are equal, and as the sides of the polygons 
forming the base are also equal, their developed sur¬ 
faces will be composed of similar and equal isosceles 
triangles, which, as we have seen (Fig. 78, a, b, c, d), 
will, when united, form a part of a regular polygon 














































SOLID GEOMETRY 


109 


inscribed in a circle, of which the inclined sides of the 
polygon form the radii. Thus in considering the base 
of the cone KH (Fig. 78) as a regular polygon of an 
infinite number of sides, its development will be found 
in the sector of a circle, MAFBM (No. 3), of which 
the radius equals the side of the cone K G (No. 1), and 
the arc equals the circumference of the circle forming 
its base (No. 2). 

To trace on the development of the covering, the 
curves of the ellipse, parabola, and hyperbola, which 
are the result of the sections of the cone by the lines 
D I, E F, I G, it is necessary to divide the circumfer¬ 
ence of the base A F B M (No. 2) into equal parts, as 
1, 2, 3, &e., and from these to draw radii to the centre 
C, which is the horizontal projection of the vertex of 
the cone j then to carry these divisions to the common 
intersection line K H, and from their terminations 
there to draw lines to the vertex G, in the vertical pro¬ 
jection No. 1. These lines cut the intersecting planes, 
forming the ellipse, parabola, and hyperbola, and by 
the aid of the intersections we obtain the horizontal 
projection of these figures in No. 2—the parabola 
passing through M E F, the hyperbola through GIL, 
and the ellipse through D I. 

To obtain points in the circumference of the ellipse 
upon the development, through the points of inter¬ 
section o, p, q, r, &c., draw lines parallel to KLI, car¬ 
rying the heights to the side of the cone G H, in the 
points 1, 2, 3. 4. 5, 6, 7, and transfer the lengths G 1, 
G 2, G 3, &c., to G 1, G 2, G 3, G 4, &c., on the radii of 
the development in No. 3; and through the points thus 
obtained draw the curve z D 1 X. 


I 


110 MODERN CARPENTRY 

• 

To obtain the parabola and hyperbola, proceed in 
the same manner, by drawing parallels to the base 
K II, through the points of intersection; and transfer¬ 
ring the lengths thus obtained on the sides of the cone 
G K, GII, to the radii in the development. 

Nos. 4 and 5 give the vertical projections of the hy¬ 
perbola and parabola respectively. 



To find the covering of the frustum of a cone, the 
section being made by a plane perpendicular to the 
axis. 

Let A C E F (Fig. 79) be the generating section of 
















SOLID GEOMETRY 


111 


the frustum. On A C describe the semicircle ABC, 
and produce the sides AE and C F to D. From the 
centre D, with the radius DC, describe the arc CH; 
and from the same centre with the radius D F, describe 


D 



the arc F G. Divide the semicircle ABC into any 
number of equal parts, and run the same divisions 
along the arc C H; draw the line H D, cutting E G in 
G; then shall CHGF be half the development of the 
covering of the frustum A C F E. 



















112 


MODERN CARPENTRY 


To find the covering of the frustum of a cone, the 
section being made by a plane not perpendicular to 
the axis. 

Let A C F E (Fig. 80) be the frustum. Proceed as 
in the last problem to find the development of the cov¬ 
ering of the semi-cone. Then—to determine the edge 
of the covering of the line EF—from the points P, q, 
r, s, t, &c., draw linos perpendicular to E F, cutting 
AC in y, x, w, v, u; and the length u t transferred 
from 1 to a, v s, transferred from 2 to b, and so on, will 
give a, b, c, d, e, &c., points on the edge of the cover¬ 
ing. 

To find the covering of the frustum of a cone, when 
cut by two cylindrical surfaces perpendicular to the 
generating section. 

Let AEFC (Fig. 81) be the given frustum, and 
A k C, E p F, the given cylindrical surfaces. Produce 
A E, CF, till they meet in the point D. Describe the 
semicircle A B C, and divide it into any number of 
equal parts, and transfer the divisions to the arc C H, 
described from D, with the radius D C. Through the 
divisions in the semicircle 1, 2, 3, 4, &c., draw lines per¬ 
pendicular to A C, and through the points where they 
intersect A C draw lines to the summit D. Draw lines 
also through the points 1, 2, 3, 4, 5, &c., of the arc C IT, 
to the summit D; then through the intersections of the 
lines from A C to D, with the seats of the cylindrical 
surfaces k, 1, m, n, o, and p, q, r, s, t, draw lines parallel 
to AC, cutting CD; and from the points of intersec¬ 
tion in C D, and from the centre D, describe arcs cut¬ 
ting the radial lines in the sector DCHin u,v w x v 

7 7 y 7 *7 y 


SOLID GEOMETRY 


113 


&c., and a, b, c, d, e, &c.; and curves traced through 
the intersections will give the form of the covering. 

VI. Spheres, Ellipsoids, &c. The development of 
the sphere, and of other surfaces of double curvature. 



is impossible, except on the supposition of their being 
composed of a great number of small faces, either 
plane, or of a simple curvature, as the cylinder and the 














114 


MODERN CARPENTRY 



cone. Thus, the sphere or spheroid may be considered 
as a polyhedron, terminated, 1st, by a greht number 
of plane faces, formed by truncated pyramids, of which 
the base is a polygon, as in Pig. 82; 2nd, by parts of 
















SOLID GEOMETRY 


115 


truncated cones forming zones, as in Fig. 83, the part 
above A B being the vertical projection, and the part 
below AB the horizontal projection; 3rd, by par.ts of 
cylinders cut in gores, forming flat sides, which dimin¬ 
ish in width, as in Fig. 84. 



Fig. 83. 


Fig. 84. 


In reducing the spheres, or spheroid, to a polyhedron 
with flat sides, two methods may be adopted, which dif¬ 
fer only in the manner of arranging the developed 
faces. 

The most simple method is by parallel circles, and 
others perpendicular to them, which cut them in two 
opposite points, as in the lines on a terrestrial globe. IT 
we suppose that these divisions, in place of being 
circles, are polygons of the same number of sides, there 
will result a polyhedron, like that represented in Fig. 
82, of which the half, A D B, shows the geometrical ele¬ 
vation, and the other half, A E B, the plan. 

To find the development, first obtain the summits 
















116 


MODERN CARPENTRY 


P, q, r, s, of the truncated pyramids, which from the 
demi-polyhedron A D B, by producing the sides Al, 
12, 23, 34, until they meet the axis E D produced; 
then from the points p, q, r, s, and with the radii P A, 
PI, q I q 2, r 2, r 3, and s 3, s 4, describe the indefinite 
arcs A B, 1 b, 1 b, 2 f, 2 f, e q, 3 g, 4 h, and from D de¬ 
scribe the arc 4 h; upon all these arcs set off the divi¬ 
sions of the demi-polygons A E B, and draw the lines to 
the summits p, q, r, s, and D, from all the points so set 
out, as A, 1, 2, 3, 4, &c., from each truncated pyramid. 
1 hese lines will represent for every band or zona the 
faces of the truncated pyramids of which they consti¬ 
tute a part. 

The development can also be made by drawing 
through the centre of each side of the polygon A E B, 
indefinite perpendiculars, and setting out upon them 
the heights of the faces in the elevation, Al 2 3 4 D, 
and through the points thus obtained drawing paral¬ 
lels to the base. On each of these parallels then set 
out the widths h, i, k, 1, d, of the corresponding faces 
(e, e, e, &c.) in the plan, and there will be thus formed 
trapeziums and triangles, as in the first development, 
but arranged differently. This method is used in con¬ 
structing geographical globes, the other is more con¬ 
venient in finding the stones of a spherical dome. 

‘ The development of the sphere by reducing it to 
conical zones (Fig. 83) is accomplished in the same 
manner as the reduction to truncated pyramids, with 
this difference, that the developments of the arrises 
indicated by A 1 2 3 4 5 B in Fig. 82, are arcs of circles 
described from the summits of cones, in place of being 
polygons. 


SOLID GEOMETRY 


117 


The development of the sphere reduced into parts 
of cylinders, cut in gores (Fig. 84), is produced by the 
second method described, but in place of joining, by 
straight lines, the points E, h, i, k, 1, d (Fig. 82), we 


G 



unite them by curves. This last method is used in 
tracing the development of caissons in spherical or 
spheroidal vaults. 

To find the covering of a segmental dome. 

In Fig. 85, No. 1 is the plan, and No. 2 the elevation 
of a segmental dome. Through the centre of the plan 

















118 


MODERN CARPENTRY 


E draw the diameter A C, and the diameter B D per¬ 
pendicular to A C, and produce B D ti I. Let D E rep¬ 
resent the base of semi-section of the dome; upon D E 
describe the arc D k with the same radius as the 
aic FGH (No. 2); divide the arc D k into any num¬ 
ber of equal parts. 1, 2, 3, 4, 5, and extend the division 
upon the right line D I, making the right line D I equal 
in length, and similar in its divisions, to the arc D k : 
from the points of division. 1, 2, 3 4, 5, in the arc D k, 
draw lines perpendicular to DE, cutting it in the 
points q, r, s, &c. Upon the circumference of the plan 
No. 1, set off the breadth of the gores or boards 1 m, 
mn, no, op, &c.; and from the points 1, m, n, o, p, 
draw right lines through the centre E: from E describe 
concentric arcs q v, ru, st, &c., and from 1 describe 
concentric arcs through the points D, 1, 2, 3, 4, 5,: 
lm, being the given breadth of the base, make lw 
equal to q v, 2x equal to ru, 3y equal to st, &c.; draw 
the curved line through the points 1, w, x, y, &c., to 1, 
which will give one edge of the board or gore to coin¬ 
cide with the line IE. The other edge being similar, 
it will be found by making the distances from the 
centre line D 1 respectively equal. The seats of the 
different boards or gores on the elevation are found 
by the perpendicular dotted lines, pp, oo, nn, m m, &c. 

To find the covering of a semicircular dome. 

Pig. 86, Nos. 1 and 2.—The procedure here is more 
simple than in the case of the segmental dome, as, the 
horizontal and vertical sections being alike, the or¬ 
dinates are obtained at once. 


SOLID GEOMETRY 


119 


To find the covering of a semicircular dome when it 
is required to cover the dome horizontally. 


G 



Fig. 86. 

Let ABC (Fig. 87) be a vertical section through the 
axis of a circular dome, and let it be required to cover 
this dome horizontally. Bisect the base in the point 
























120 


MODERN CARPENTRY 


D, and draw D B E perpendicular to A C, cutting the 
circumference in B. Now divide the arc B C into 
equal parts, so that each part will be rather less than 



the width of the a board; and join the points division 
by straight lines, which will form an inscribed polygon 
of so many sides; and through these points draw line 
parallel to the base A C, meeting the opposite sides of 

































SOLID GEOMETRY 


121 


the circumference. The trapezoids formed by the sides 
of the polygon and the horizontal lines may then be 
regarded as the sections of so many frustums of cones; 
whence results the following mode of procedure, in 



Fig. 88 . 

accordance with the introductory illustration Fig. 82; 
—Produce, until they meet the line D E, the lines g f, 
f, n, &c., forming the sides of the polygon. Then, to de¬ 
scribe a board which corresponds to the surface of one 
of the zones, as f g, of which the trapezoid m 1 f g is a 


















122 


MODERN CARPENTRY 


section,—from the point h, where the line f g produced 
meets D E, with the radii li f, hg, describe two arcs, 
and cut off the end of the board k on the line of a 
radius h k. 

To obtain the true length of the board, proceed as in 
Fig. 89. The other boards are described in the same 
manner. 



To find the covering of an elliptic dome. 

Let ABCD (Fig. 90) be the plan, and FGH the 
elevation of the dome. Divide the elliptical quadrant 
F G (No. 2) into any number of equal parts in 1, 2, 3, 
4, 5, and draw through the points of division lines 

















SOLID GEOMETRY 


123 


perpendicular to FH, and produced to A C (No. 1), 
meeting it in i, k, m, n ,: these divisions are transferred 
, by the dotted arcs to the gore b E c and the remainder 
of the process is as in Figs. 85 and 86. 

To find the covering boards of an ellipsoidal dome. 

Let A B C D (No. 1, Fig. 89) be the plan of the dome, 
and FGH (No. 2) the vertical section through its 
major axis. Produce FH indefinitely to n; divide the 
circumference, as before, into any number of equai 
parts, and join the divisions by straight lines, as p m, 
ml, &c. Then, describe on a board, produce the line 
forming one of the sides of the polygon, such as 1 m, to 
meet F n in n; and from n, with the radii n m, n 1 
describe two arcs forming the sides of the board, and 
cut off the board on the line of the radius n o. Lines 
drawn through the points of divisions at right angles 
to the axis, until they meet the circumference ADC 
of the plan, will give the plan of the boarding. 

To find the covering of an ellipsoidal dome in gores. 

Let the ellipse ABCI) (Fig. 90, No. 1) be the plan 
of the dome, A C its major axis and B D its minor axis; 
and let ABC (No. 2) be its elevation. Then, first, to 
describe on the plan and elevation the lines of the 
gores, proceed thus:—Through the line AC (No. 1) 
produced at II, draw the line E G perpendicular to it, 
and draw BE, D G, parallel to the axis A C, cutting 
E G; then will E G be the length of the axis minor, 
on which is to be described the semi-circle E F G, rep¬ 
resenting a section of the dome on a vertical plane 
passing through the axis minor. 

Divide the circumference of the semi-circle into any 
number of equal parts, representing the widths of the 


124 


MODERN CARPENTRY 


covering boards on the line B D; and through the 
points of division 1. 2, 3, 4, 5, draw lines parallel to the 
axis A C, cutting the line B D in 1, 2, 3, 4, 5. Divide 
the quadrant of the ellipse C D (No. 1) into any num¬ 
ber of equal parts in e, f, g, h, and through these points 
draw the lines e a, f b, g c, h d in both diagrams, per- 




Fig. 90. 


pendicular to A C, and these lines will then be the seats 
of vertical sections through the dome, parallel to 
EFG. Through the points e, f, g, h (No. 1) draw 
lines parallel to the axis A C, cutting E G in o, n, m, k; 













































































SOLID GEOMETRY 


125 


and from H, with the radii Ho, H n, H m, H k, de¬ 
scribe concentric circles o 9 9 p, n 8 8q, m z y r, &c. To 
find the diminished width of each gore at the sections 
a, e, b f, c g, d h:—Through the divisions of the semi¬ 
circle, 1, 2, 3, 4, 5, draw the radii Is, 2t, 3u, 4v, 5 w, 6x; 
then by drawing through the intersections of these 
radii with the concentric circles, lines parallel to HF, 
to meet the section lines corresponding to the circles, 
the width of the gores at each section will be obtained; 
and curves through these points will give the repre¬ 
sentation of the lines of the gores of the plan. 

In No. 2 the intersections of the lines are more 
clearly shown. The quadrant E G F is half the end 
elevation of the dome, and is divided as in No. 1. The 
parallel lines 5 5, 4 4, 3 3, show how the divisions of 
the arc of the quadrant are transferred to the line D B, 
and the other parallels ah, b k, cl, d m, are drawn 
from the divisions in the circumference of the ellipse 
to the line E G, and give the radii of the arcs m, 1 o, 
k p, h q. 

To describe one of the gores draw any line AB 
(No. 3), and make it equal in length to the circumfer¬ 
ence of the semi-ellipse A D C, by setting out on it the 
divisions 1, 2, 3, 4, 5, &c., corresponding to the divisions 
C h, h g, g f, &c., of the ellipse: draw through those 
divisions lines perpendicular to A B. Then from the 
semi-circle (No. 1) transfer to these perpendiculars the 
widths 6 5 to g n, 9 9 to f m, 8 8 to e 1, y z to d k, and 
x w to c h, and join Ac, d d, de, e f, f g, A h, h k, k 1, 
1 m, and m n,; which will give the boundary lines of 
one-half of the gore, and the other half is obtained in 
the same manner. 


126 


MODERN CARPENTRY 


To describe the covering of an ellipsoidal dome with 
horizontal boards of equal width. 

Let A B C D (No. 1, Fig. 91) be the plan of the dome, 
ABC (No. 2) the section on its major axis, and L M N 
the section on its minor axis. Draw the circumscribing 
parallelogram of the ellipse, namely, FGHK (No. 1), 
and its diagonals F IIG K. In No. 2 divide the cir¬ 
cumference into equal parts, 1, 2, 3, 4, representing the 
number of covering boards, and through the points of 
division draw lines 18, 2 7, &c., parallel to A C. 
Through the points of divisions draw 1 p, 2 t, 3 x, &c., 
perpendicular to A C, cutting the diagonals of the cir¬ 
cumscribing parallelogram of the ellipse (No. 1), 
and meeting its major axis in p, t, x, &c. Complete 
the parallelograms, and inscribe ellipses therein cor¬ 
responding to the lines of the covering. Produce the 
sides of the parallelograms to intersect the circumfer¬ 
ence of the section on the minor axis of the ellipse in 
1, 2, 3, 4, and lines drawn through these parallel to 
L N, will, give the representation of the covering boards 
in that section, lo find the development of the cover- 
ing, produce the axis D B, in No. 2, indefinitely. Join 
by a straight line the divisions 1 2 in the circumfer¬ 
ence, and produce the line indefinitely, making e k 
equal to e 2, and k g equal to 1 2; 1 2e k g will be the 
axis major of the ellipses of the covering 12 7 8. Join 
also the corresponding divisions in the circumference 
of the section on the minor axis, and produce the line 
12b to meet the axis produced; and the length of 
tins line will be the semi-axis minor, eh, of the ellipse, 

2h k, while the width f h will be equal to the division 
12 in NM L. 


I 


SOLID GEOMETRY 12? 



Fig. 91. 
















































128 


MODERN CARPENTRY 


To find the covering’ of an annular vault. 

Let A C K G E F A (Fig. 92) be the generating sec¬ 
tion of the vault. On A C describe a semi-circle ABC, 
and divide its circumference into equal parts, repre¬ 
senting the boards of the covering. From the divisions 
of the semi-circle, b, m, t, &c., from the centre D of 



the annulus, with the radii D r, D s, &c., describe the 
concentric circles, s q, &c., representing the covering 
boards in plan. Through the centre D draw II K per¬ 
pendicular to G C, indefinitely extending it through K. 
Join the points of division of the semi-circle, A b, b m, 
m t, by straight lines, and produce them until they cut 






















SOLID GEOMETRY 


129 


the line K H as mbn, t m u, when the points n, u, &c., 
are the centres from which the curves of the covering 
boards mo, tv, &c., are described. 



To find the covering of an ogee dome, hexagonal in 
plan. 

Let A B C D E F (No. 1, Fig. 93) be the plan of the 
dome, and II K'L (No. 2) the elevation, on the di¬ 
ameter F C. Divide H K into any number of equal 
parts in 1, 2, 3, 4, 5, k, and through these draw per¬ 
pendiculars to H L, and produce them to meet F C 




























130 


MODERN CARPENTRY 


(No. 1) in 1, mm, n, o, p, G. Through these points 
draw lines 1 d, me, n f, &c.; parallel to the side F E of 
the hexagon: bisect the side F E in N, and draw G N, 
which will be the seat of a section of the dome, at right 
angles to the side E F. To find this section nothing 
more is required than to set up on N G from the points 
t, u, v, &c., the heights of the corresponding ordinates 
q 1, r 2, s 3, &c., of the elevation (No. 2) to draw the 
ogee curve N 1 2 3 4 5 p, and then to use the divisions 
in this curve to form the gore or covering of one side 
E g h k M D. 


PART II. 

PRACTICAL SOLUTIONS. 

Having taken a thorough course in Solid Geometry, 
the student should be now prepared to solve almost 
any problem in practical construction, almost as soon 
as they present themselves. The various problems in 
construction, however, are so numerous, and in many 
cases, so intricate, that the student will often be con¬ 
fronted by problems which will require so many appli¬ 
cations of the rules he has been taught, in different 
forms, that without some helping guidance, he will fail 
to see exactly what to do. 

The following examples, with their explanations, are 
intended to give him the aid necessary to solve many 
difficult problems, and equip him with the means of still 
further investigation and sure results. 

It is often necessary for the workman to find the 
exact stretchout or length of a straight line that shall 
equal the quadrant or a semi-circle. 

To accomplish this:—Take A B radius, and A cen¬ 
tre; intersect the circle at C; join it and B; draw 
from D, parallel with C B, cutting at II; then A II will 
be found equal to curve A D. Pig. 1. 

This method is somewhat different to that already 
given; both, however, are practical. 

Fig. 2. To find a straight line which is equal to the 
circumference of a circle. Draw from the centre, 0, 

131 


MODERN CARPENTRY 













PRACTICAL SOLUTIONS 


133 


any right angle, cutting at J and V; join J Y; draw 
from 0 parallel with J V; square down from J, cut¬ 
ting at N; join it and V; then four times NY will be 
found to equal the circumference. 

How to find the mitres for intersecting straight and 
circular mouldings. 

Figure 3 shows the form of an irregular piece of 
framing or other work, which requires to have mould¬ 
ings mitre and properly intersect. 

The usual way of doing this is to bisect each angle, 
or to lay two piecs of moulding against the sides of 
framing, and mark along the edge of each piece, thus 
making an intersectic or point, so that by drawing 
through it to the next point, which is the angle of 
framing, the direction of mitre is obtained. This pro¬ 
cess, however, is not the quickest and best by any 
means. The most simple and correct method is to ex¬ 
tend the sides A L and P H. 

Now suppose we wish to find a mitre from L; take 
it as centre, and with any radius, as K, draw the circle, 
cutting at J; join it and K; draw from L parallel with 
J K, and we have the mitre at once. 

Now come to angle on the right; here take II as 
centre, and with any radius, as E, draw the circle, 
cutting at F; join it and Ej draw from LI parallel with 
E F, and you will find a correct mitre. 

The next question is the intersection of straight and 
circular mouldings. 

In the present case an extreme curve is given, in 
order to show the direction of mitre here, which is 
simply on the principle of finding a centre, for three 
points not on a straight line. For example, ABC are 


134 


MODERN CARPENTRY 







*- 


Fig. 3. 










PRACTICAL SOLUTIONS 


135 


points; bisect A B and B C; draw through intersec¬ 
tions thus made, and lines meeting in point D give a 
centre, from which strike the circular mitre as shown. 

Here it may be stated that in some cases a straight 
line for mitres will answer; this means when the curve 
is a quadrant or less. 

Fig. 4 shows the intersections of rake and level 
mouldings for pediments. 

The moulding on the rake, increases in width, and 
is entirely different from that on the. level, yet both 
mitre ,and intersect, the rake moulding being worked 
to suit the level. If the curves of Fig. 4 are struck from 
centres as shown, then by the same rule, the rake 
moulding is also struck from centres. 

Take any point in the curve, as C; square up from it, 
cutting at B; draw from C parallel to SL; join L K, 
which bisect at N; make E D equal to A B on the right; 
join L D and D N; bisect L D, also D N; draw through 
intersections thus made, and the lines meeting in F, 
give a point from which draw through N; make N J 
equal H F; then F and J are centres, from which strike 
the curve, and it will be found to exactly intersect 
with that of Fig. 4. 

Both mouldings here are shown as solid, and of the 
same thickness. This is done for the purpose of mak¬ 
ing the drawing more plain and easily understood; 
but bear in mind that all crown mouldings are gener¬ 
ally sprung. 

To find the form of a sprung or solid moulding on any 
rake without the use of either ordinates or centres. 

It may not be generally known, that if a level mould¬ 
ing is cut to a mitre, that the extreme parts of mitre ? 


136 


MODERN CARPENTRY 





Fig. 4. 













PRACTICAL SOLUTIONS 


137 


when in a certain position, will instantly give the 
exact form of a rake moulding, and it will intersect, 
and mitre correctly with that of level moulding. To 
do this, take the level piece which has been mitred; 
lay its flat surface on the drawing; make its point P 
at Pig. 5, stand opposite point P at Fig. 6; keep the 
outer edge fair with line N L. The piece being in this 
position, take a marker, hold it plumb against the 
mitre, and in this way, prick off any number of points 
as shown, through which trace the curve line, and the 
result is a correct pattern by which the rake moulding 
is worked. 

A moment’s consideration will convince us that this 
simple method must give the exact form of any rake 
moulding to intersect with one on the level. 

To cut the mitres and dispense with the use of a 
box, this method wull be found quick and off-hand. 
Take, for example, the back level moulding, and square 
over on its top edge any line, as that of FN; con¬ 
tinue it across the back to H; make HY equal TL 
above, and from Y, square over lower edge H K. Now 
take bevel 2 from above, and apply it on top edge, 
as shown; mark F L; then join L Y; cut through these 
lines from the back, and the mitre is complete. 

To cut the mitre on the rake moulding, square over 
any line on its back, as that of H J; continue it across 
the top and lower edge; take bevel X, shown above 
Fig. 5, apply it here on top edge, and mark D A; take 
the same bevel, and apply it on small square at E, and 
mark E 2. 

We now want the plumb cut on lower edge J K, and 
the same cut on front edge N P, shown at Fig. 6. Take 


138 MODERN CARPENTRY 



Fig. 5. 


Fig. Q. 




























PRACTICAL SOLUTIONS 


139 


bevel W above Fig. 5, apply it here and mark 2B; join 
B A: this done, apply the same bevel on front edge 
N P, and mark the plumb cut, it being parallel with 
that of 2B here, or K J, Fig. 5; now cut through lines 
on the back, and the mitre is complete. 

It has already been shoAvn, that we dispense with 
making or using a box for mitreing sprung mouldings. 

In this case, the front edge or upper member, stands 
parallel with face of wall, so that bevel X being ap¬ 
plied, gives the plumb cut; then the cut on top edge 
is square with face of wall. This shows, that we have 
only to find the direction of a cut on the back of mould¬ 
ing to make the mitre. 

To do this, take any point as R; draw from it square 
with rake of gable. Now mark sections of moulding, 
as shown, its back parallel with R F; draw from D 
square with E N; extend the rake to cut line from D at 
K; this done, take any point on the rake, say L; draw 
from it parallel with R F, cutting at K; take it as 
centre and L as radius, and draw the arc of a circle; 
with same radius return to K on the right; take it as 
centre, and draw the arc L Hj make the first arc equal 
it; then draw from H parallel with L C, cutting at 
C J; draw from it square with rake, cutting at C, and 
join C K. This gives bevel W for cut on back of mould- 
ing. 

A most perfect illustration of this may be had by 
having the drawing on card-board, and cutting it clear 
through all the outer lines, including that of the mould¬ 
ing on lines FDNE, making a hinge by a slight cut 
on line R F; also make a hinge of line R A, by a slight 
cut on the back, and in like manner make front edge 


140 MODERN CARPENTRY 

work on a hinge by a slight cut on line F Y. This cut 
is made on top surface. Perform the same operation 

,5 



on the left. All the cuts being made, raise both sides 
on hinges A R and A 2; push the sections of mould¬ 
ings on right and left from you; make front edge rest 










141 


PRACTICAL SOLUTIONS 

on F D. Now bring mitres together, and we have a 
practical illustration of mitreing sprung mouldings on 
the rake. (See Fig. 7.) 


PROJECTION OF SOLIDS. 

The following illustrations will.be found by the stu¬ 
dent almost indispensable in the construction of vari¬ 
ous objects, and they will open the door to many more. 


Jf c x 



Fig. 8. 


Fig. 8 shows the projection of a solid. This means 
the section of anything that is cut by a plane not paral¬ 
lel to its base; or, to put this in a more practical way 
—take a square bar of wood and cut it in the direction 
of BE; the section it makes is shown by B E FII; 
simple as this is, it still gives the idea of what is meant 
by projection. 






142 MODERN CARPENTRY 

Fig. 9 shows the section of a square bar which has 
been cut by two unequal pitches, say in the direction 
of bevels J and H; the line C B is called the seat; 
from it all measurements are taken and transferred to 
lines that are square with the pitch AB; this pitch 
may be called a. diameter, because it and the ordinate 
A E are at right angles. 


JS 



Fig. 10 shows the sides of a square bar which may 
be any length. The bar is to stand perpendicular, and 
pass through a plank that inclines at AB; the learner 








PRACTICAL SOLUTIONS 143 

is now required to show on the surface of plank the 
shape of a mortise that shall exactly fit the bar. 



To solve this the student is left to exercise his own 
intelligence. 

The problem may be clearly demonstrated by a 
card-board model; it being cut, and the parts pro- 







144 


MODERN CARPENTRY 


jected from the flat surface will represent the plank, 
and show the mortise in it standing directly over the 
square, Fig. 10. 

Fig. 11. It is here required to mark two unequal 
pitches on two sides of a square bar—then to have a 
piece of plank or board cut so that it shall exactly fit 
both pitches on the bar. The inclination of plank may 
be assumed as A B, and height of both pitches as II A; 
let E F be the sent from which all measurements are 
taken, and transferred to lines that are on the surfaces 
of plank and square with AB; thus giving points to 
direct in drawing line CD; and DE produced. 

Could we apply the bevel J to points C and K, and 
have plumb lines on the edge of plank, then by cutting 
through these and those already marked on the sur¬ 
face, the problem would at once be solved, by making 
both upper and under surface of plank fit the two 
pitches as required. But in practice this would be in¬ 
admissible on account of the great waste of material. 

The proper method is to take any point, E, and cut 
through the plank square with E D, and at C, cut 
through the plank square with C D; here it will be 
noticed that line A D on the surface makes a very 
different pitch to that of A B on the edge of plank, and 
that C D differs from both. 

To understand these points thoroughly is the true 
secret of the nicest element in the joiner’s art—hand 
railing. It being clear that two bevels are required 
one for each pitch—proceed to find them. Make N L 
equal one side of the square. Take N as centre, and 
strike an arc touching the line E D; with same radius, 
and L centre, make the intersection in S; join it at L, 


PRACTICAL SOLUTIONS 145 



Fig. 11. 














146 


MODERN CARPENTRY 


which gives bevel R for joint at E. Again take N as 
centre, and strike an arc touching line D C; with same 
radius, and L centre, make the intersection in W; join 
it and L, which gives bevel P for joint at C; the cuts 
being made by these bevels. 


PANELLED CEILINGS IN WOOD AND STUCCO 
BRACKETS, AND SIMILAR WORK. 

It is no part of the duty of this work to show designs 
for wooden ceilings, but in order to illustrate the 
method or methods, of constructing a wooden ceiling 
I deem it proper to show a design in this style the better 
to convey to the student the reason for the various 
steps taken to reach the desired result. 

Fig. 12 shows a section of a roof and the mode of 
constructing a bracketed ceiling under it, but with 
slight modification the same arrangement can be 
adopted for ceilings under floors. The rafters A A are 
18 inches from centre to centre. On these, straps a a, 
3X1^ inches, are nailed at 16 inches apart, and similar 
straps a a, IV 4 XI inch, are nailed to the ceiling joists. 
To the straps are nailed the brackets b b b, shaped to 
the general lines of the intended ceiling, and also 
placed 14 inches apart. The laths are nailed to the 
brackets for the moulded parts, and to straps for the 
flat parts of the ceiling. The brackets and the straps 
for ceiling should not be more than one inch in thick¬ 
ness, for where the brackets and straps occur the 
plaster cannot be pressed between the laths to form a 
key. If the brackets and straps are made thicker, 


PRACTICAL SOLUTIONS 


147 












































































































148 


MODERN CARPENTRY 


parts of the ceiling are apt to be weak and irregular. 
The lathing should be well bonded and have no end 
joints longer than twelve inches on the ceiling, and 
twenty-four inches on the wails and partitions. No 
joints of laths should be overlapped, as the plaster 
would thereby be made thinner at a part where it 
forms a no key, and would thus be liable to crack from 
the vibration of roof, floor or other causes. 

Fig. 13 is a section on a larger scale at right angles 
to that shown in Fig. 12. Here the ceiling bracket b 
is shown affixed by hooks to the wall, and to the ceiling- 
joists by the strap a a. The ceiling having been plas¬ 
tered, and the mouldings of the cornice run on the 
lathed brackets prepared to receive them, the plaster 
enrichments marked c d e f g are then applied. No. 3 
on Fig. 2 shows a curtain-box in section with its cur¬ 
tain-rod. 

Fig. 14 is a section through a window-head, meeting- 
rails and sill of a window. The safe lintel is placed 
about 10 inches above the daylight of the window, for 
the purpose of allowing Venetian blinds to be drawn up 
clear of the window, and leave the light unobstructed. 
The framing of the window is carried up to the lintel, 
and between it and the upper sash a panel is set in, 
and the blinds hang in front of it. 

Fig. 15 is a cross vertical section, and Fig. 16 a plan, 
looking up, of a skylight on the ridge of a roof, suitable 
for a staircase or corridor. The design can be adopted 
to suit various widths. The skylight is bracketed for 
plaster finishings, in the same manner as the ceiling 
already described. The framing and mouldings at b 
are carried down the side of the light at the same slope 




PRACTICAL SOLUTIONS 





















































































































150 


MODERN CARPENTRY 


as the sash, till they butt against the sill and bridle 
d and e, forming a triangular panel having for its base 
the cornice c, which is carried round the aperture 
horizontally, and finishing flush with the ceiling, per¬ 
mits the cornice on the corridor to be continued with- 



Fig. 16. 


out interruption. Observe in Fig. 16 the lower cornice 
finishes the walls, and the upper mouldings 0 marked 
c in Fig. 15 are carried round the well of the skvlijrht. 

b lg. 17 shows a plan of a panelled ceiling. The lower 
members or bed mouldings are carried round the walls 
of the room, and tend to buil 1 together, and give an 




































































PRACTICAL SOLUTIONS 


151 


appearance of support to the several parts of the ceil¬ 
ing. The best way of making panelled ceilings is to 
cover the floor with boarding, and lay down the lines 
of the ceiling on- the temporary floor thus formed. 
Then build and lath the ceiling on these lines. When 
it is completed it will be quite firm, and can be cut into 
sections suitable for being lifted up and attached to 
the joisting. The same lines will serve as guides for 
the plasterer setting the moulds for running the cor¬ 
nices and for preparing the circular ornament. 



Fig. 17. 


Fig. 18 shows in plan and section a centre suited for 
this ceiling. By fixing the central portion of it one 
inch from the surface of the finished ceiling and con¬ 
necting it by small plaster blocks placed about one 
inch apart, it forms an excellent ventilator. A zinc 
tube can be led from this centre into a vent, and an 




















































































































































152 


MODERN CARPENTRY 


air tight valve put on the tube to prevent a down draft 
when the vent is not in use. F is an iron rod fixed to 




% 


a strut or dwang between the joists for the purpose 
of securing a gas pendant or electrolier. 

Let CAB (Fig. 19) be the elevation or the bracket 
of a core, to find the angle-bracket. 








































































I 


PRACTICAL SOLUTIONS 153 


First, when it is a mitre-bracket in an interior angle, 
the angle being 45° : divide the curve C B into any 






























154 


MODERN CARPENTRY 


A B, and cutting it in d e f g c; and produce them to 
meet the line D E, representing the centre of the seat 
of the angle-bracket: and from the points of intersec¬ 
tion h i k 1 c draw lines h 1, i 2, k 3, 14, at right angles 
to D E, and make them equal—h 1 to d 1, i 2 to e 2, 
&c.; and through F 1 2 3 4 5 draw the curve of the 
edge on the bracket. .The dotted lines on each side 
of D E on the plan show the thickness of the bracket, 
and the dotted lines ur, vs w t, show the manner of 
finding the bevel of the face. In the same figure is 
shown the manner of finding the bracket for an obtuse 
exterior angle. Let D I K be the exterior angle: bisect 
it by the line I G, which will represent the seat of the 
centre of the bracket. The lines I H, m 1, n 2, o 3, p 4, 
c 5, are drawn perpendicular to I G, and their lengths 
are found as in the former case. 

To find the angle-bracket of a cornice for interior 
and exterior, otherwise reentrant and salient, angles. 

Let AAA (Fig. 20) be the elevation of the cornice- 
bracket, E B the seat of the mitre-bracket of the in¬ 
terior angle, and II G that of the mitre-bracket of the 
exterior angle. From the points A k a b c d A, or 
wherever a change in the form of the contour of the 
bracket occurs, draw lines perpendicular to Ai or 
D C, cutting A i in e f g h i and cutting the line E B 
in E 1 m n o B. Draw the lines E G, G L B H, and H K, 
representing the plan of the bracketing, and the par¬ 
allel lines from the intersection lmno, as shown 
dotted in the engraving; then make B F and III jper- 
pendicular to E B and GII respectively, and each 
equal to i A, o u to h d, n t to g c, m s to f b, 1 r to e a, 

1 P to e k, and join the points so found to give the con- 


PRACTICAL SOLUTIONS 


155 


tours of the brackets required. The bevels of the face 
are found as shown by the dotted lines x v y w, &c. 



To find the angle-bracket at the meeting of a concave 
wall with a straight wall. 

Let ADEB (Fig. 21) be the plan of the bracketing 
on the straight wall, and DMGE the plan on the cir- 































156 


MODERN CARPENTRY 


cular wall, CAB the elevation on the straight wall, 
and G M H on the circular wall. Divide the curves 



Angle-bracket at Meeting of Straight 
and Concave Walls 


Fig. 21. 


C B, G II into the same number of equal parts; through 
the divisions of C B draw the lines C D, 1 d h, 2 e i, &c., 
perpendicular to A B and through those of G H draw 





















PRACTICAL SOLUTIONS 


157 


the parellel lines, part straight and part curved 1 m h, 
2 n i, 3 o k, &c. Then through the ^intersections h i k 1 
of the straight end curved lines draw the curve D E, 
which will give the line from which to measure the 
ordinates h 1, i 2, k 3, &c. 



Fig. 22 shows the method of finding the angle-bracket 
at the meeting of coves of equal height but unequal 
projection. The height C B is equal to G H, but the 
projection B A is greater than H I. 































158 


MODERN CARPENTRY 


Fig. 23, Nos. 1, 2, 3, shows the curb and ribs of a 
circular opening (CBA, No. 2), cutting in on a slop¬ 
ing ceiling. No. 1 is a section through the centre B D, 
No. 2 and E F I, No. 3. The height L K is divided into 



Fig. 23. 


equal parts in e, f, g, h, i, and the same heights are 
transferred to the main rib in No. 2 at A, 1, 2, 3, 4, 5, B. 
Through the points A, 1, 2, 3, 4, 5, in No. 2 lines are 
drawn parallel to the axis BEI; and through the 
points e, f, g, h, i. in No. 1 lines are drawn parallel to 












































































PRACTICAL SOLUTIONS 


159 


the slope I(H. The places of the ribs 1, 2, 3, 4, 5, in 
the latter, and their site on the plan, No. 3, and also 
the curve of the curb, are found by intersecting lines 
in the manner with which the student is already ac¬ 
quainted. 



be 



N°3 


Spherical Niches on Semicircular 
and Segmental Plans 


Fig. 24. 


ON NICHES. 

To describe a spherical niche on a semi-circular plan. 

The construction of this (Fig. 24) is precisely like 
that of a spherical dome. The ribs stand in planes, 
which would pass through the axis if produced. They 
are all of similar curvature. No. 2 shows an elevation 
















































160 


MODERN CARPENTRY 


of the niche, and No. 3 the bevelling of the ribs a, b, 
against the front rib at D on the plan; a b is the bevel 
of a, and b c of b. 

Let HBK (Fig. 24) be the plan. It is obvious that 
the ribs m p, nr, will be parts of the quadrant G F 
(No. 3). Transfer the lengths lo, m p, nr, and r s to 
the line G E, as shown at opr s, and raise perpendicu¬ 
lars from these points to the quadrant; G p is the rib 
m p, and o p is the bevel; G r is the rib nr, and r s is 
the section of the front rib at the crown; the vertical 
projection of the upper arris of this rib will be a semi¬ 
circle with radius s s or s H. 

The niche of which both the plan and elevation are 
segments of a circle. 

No. 25 is the elevation of the niche, being the seg¬ 
ment of a circle whose centre is at E. No. 1, A B C, is 
the plan, which is a segment of a circle whose centre 
is I). Having drawn on the plan as many ribs as are 
required, radiating to centre D, and cutting the plan 
of the front rib in a, b c, de; then through the centre 
D draw the line G II parallel to A C; and from D de¬ 
scribe the curves m 1, AG, C II, cutting the line GH; 
and make D F equal to E 0, No. 2. From F as a centre 
describe the curves 1 p 1 and G I H for the depth of 
the ribs; and this is the true curve for all the back 
ribs. 

To find the lengths and bevel of the ribs:—From the 
centre D describe the quadrant and arcs a f, b g, c g, 
d h, &c., and draw f f, g g, li h perpendicular to D II, 
cutting the curve 1 p 1, and the lines of intersection will 
give the lengths and bevels of the several ribs. 


PRACTICAL SOLUTIONS 


161 



Fig. 25. 



































































MODERN CARPENTRY 



G 

Elliptical Niche on Seg.oental Plan 

Fig. 26. 














































































PRACTICAL SOLUTIONS 


163 


Let D in the plan (No. 1, Fig. 26) be the centre of 
the segment. Through D draw E F parallel to A C, 
and continue the curve of the segment to E F. Then 
to find the curve of the back ribs:—From klmn, any 
points in the curve of the front rib (No. 2) let fall per¬ 
pendicular to the line A B, cutting it in a b c d. Then 
from D as a centre describe the curves a e, b f, c g, e h, 
d h, and from the points where they meet the line E F 
draw the perpendiculars e k, fl, gm, hn, ho, and set 
up on e k the height ek of the elevation and the cor¬ 
responding heights on their other ordinates, when 
klmno will be the points through which the curve 
of the radial ribs may be traced. The manner of find¬ 
ing the lengths and bevels of the ribs is shown at 
t u u v v. 

A niche semi-elliptic in plan and elevation. 

Let No. 1, Fig. 27, be the plan, and No. 2 the eleva¬ 
tion of the niche. The ribs in this case radiate from the 
centre D, and with the exception of m g (which will 
be the quadrant of a circle) they are all portions of 
ellipses, and may be drawn by the trammel, as shown 
in No. 4, which gives the true curve of the rib marked 
d in No. 2 and b D in No. 1. The rib c, in the eleva¬ 
tion, is seen at a D in No. 1; the bevel of the end h i is 
seen at A a in No. 3, and that of the end e f at b c. 

To draw the ribs of a regular octagonal niche. 

Fjg. 28.—Let No. 1 be the plan, and No. 2 the eleva¬ 
tion of the niche. It is obvious that the curve of the 
centre rib H G will be the same as that of either half 
of the front rib A G, F G. In No. 3, therefore, draw 
ABODE, the half-plan of the niche, equal to 
A B C II G, No. 1, and make D G E equal to half the 


MODERN CARPENTRY 


164 






6 3 ±4 


Him 


1 

=fc 


£ 


*n 


Elliptical Niche on Elliptical Plan 
Fig. 27. 






































































PRACTICAL SOLUTIONS 


165 



front rib. Divide D G into any number of parts 12 3 4, 
&c., and through the points of division draw lines par¬ 
allel to A G, meeting the seat of the centre of the angle 











































































MODERN CARPENTRY 




-Irregular Octagonal Niche 

Fig. 29, 
























































































PRACTICAL SOLUTIONS 


167 


rib C E in i k 1 m n o. On these points raise indefinite 
perpendiculars, and set up on them the heights a 1 in 
i 1, b 2 in k 2, and so on. The shaded parts show the 
bevel at the meeting of the ribs at G in No. 1. 

To draw the ribs of an irregular octagonal niche. 

Fig. 29.—Let No. 1 be the plan, and No. 2 the eleva¬ 
tion of the niche. Draw the outline of the plan of the 
niche at ABCDEF (No. 3), and draw the centre 
lines of the seats of the ribs B G, H I, &c.; draw also 
GLF equal to the half of the front rib, as given in the 
elevation No. 2, and divide it into any number of parts 
12 3 4. Through the points of division draw dl, c 2, 
b 3, a 4, perpendicular to G F, and produced to the 
seat of the first angle rib G E. Through the points of 
intersection draw lines parallel to the side E D of the 
niche meeting the second angle rib DG; through the 
points of intersection again draw parallels to D C, and 
so on. The curve of the centre rib is found by setting 
up from n o p q G the heights d 1, c 2, &c., on the par¬ 
allel lines which are perpendicular to K G. The curve 
of the rib B G or E G is found by drawing through the 
points of intersection of the parallels perpendiculars 
to the seat of the rib, and setting upon them, at 
h m r I G, the heights dl, c2, &c. No. 4 shows the rib 
C G, and No. 5 the intermediate rib TII. 

DOUBLE CURVATURE WORK. 

To Obtain to Soffit Mould for marking the veneer 
(see Fig. 35), divide the elevation of lower edge of the 
head (Fig. 30) into a number of equal parts, as A, B, 
C, D, E, S, and drop projectors from these points into 


168 MODERN CARPENTRY 

the plan cutting the chord line A" S" in A", B", C", 
D", E", S". Draw the line s' s', Fig. 35, equal in length 
to the stretch out of the soffit in the elevation (the 
length of a curved line is transferred to a straight 
one by taking a series of small steps around it with 
the compasses, and repeating a like number of the 
straight line), and transfer the points ABC, &c., as 
they occur thereon, repeating them on each side of the 
centre line. Erect perpendiculars at the points, and 
make each of these lines equal in length to its cor- 
despondingly marked line in the plan, as A' a' a' Fig. 
35, equal to A" a a Fig. 32, these letters referring re¬ 
spectively to the chord line and the inside and outside 
edges of the head. Draw the curves through the points 
so found. As will be seen by reference to Fig. 35, the 
mould is wider at the springing than at the crown; 
this is in consequence of the pulley stiles being parallel. 
If they were radial their width would be the same as 
the width of the head at the crown, and the head would 
be parallel; the gradual increase in width from the 
crown to the springing is also apparent in the sash- 
head and the beads, as indicated by the line 0 0. Fig. 
35, which is the inside of the sash-head, and the out¬ 
side of the parting head; this variation in width ren¬ 
ders it impossible to gauge to a thickness from the face 
or the groove in the head from its edges. 

To Form the Head. Having prepared the cylinder 
(Fig. 33) to the correct size, prepare a number of 
staves to the required section, which may be obtained 
by drawing one or two full size on the elevation, as 
shown to the right in Fig. 36. The staves should be 
dry straight-grained yellow deal, free from knots and 


PRACTICAL SOLUTIONS 


169 


Fig. 30. 


Fig. 31. 



Fig. 32. 



















































































































































































170 MODERN CARPENTRY 



Fig. 35. 


sap, and not be so wide that they 
require hollowing to fit. If the 
veneer is pine, it will probably bend 
dry, but hardwood will require soft¬ 
ening with hot water. One end 



should be fixed as shown in Fig. 34, 
by screwing down a stave across it. 
Then the other end is bent gently 
over until the crown is reached, 


¥ 

* 

=4 

— I- 

i • 

# * 

• • 

• • 

1 • 

» • 

r 

b— - , 1 


- 


1 TJ -- J ' 

f zJ 


— -m. —* 1 


-- 

-•H 

^i 

F. 





Fig. 34. 


when another stave is screwed on, 
and the bending continued until 
the veneer is well down all round, 
and a third stave secures it until 
it is thoroughly dry, when the re^ 














































PRACTICAL SOLUTIONS 


171 


mainder may be glued on. It is as well to interpose a 
sheet of paper between the cylinder and the veneer, in 
case any glue should run under, which would then ad¬ 
here to the paper instead of the veneer. The head 
should not be worked for at least twelve hours after 
glueing. If a band saw is at hand, the back should be 
roughly cleaned off and the mould bent round it, the 
shape marked, and the edges can then be cut vertically 
with the saw, by sliding it over the cylinder sufficiently 



for the saw to pass. When cut by hand, the mould 
is applied inside and the cut is made square to the face, 
the proper bevel being obtained with the spokeshave, 
and found by standing the head over its plan and 
trying a set square against it. When fitting the head 
to the pulley stiles the correctness of the joints is tested 
by cutting a board to the same sweep as the sill, with 









































172 


MODERN CARPENTRY 


tenons at each end. and inserting it in the mortises for 
the pulleys, as shown at B, Fig. 36. A straight-edg* 
applied to this and the sill will at once show if the 
head is in the correct position, and if the edges are 
vertical as they should be. 

To Obtain a Developed Face Mould. Make the line 
i li, Fig. 37, equal to the stretch out of the plan of the 
face of sash-heads, viz., I H, Fig. 32. Transfer the di¬ 
visions as they occur, and erect perpendiculars thereon. 



Fig. 37. 


Make these equal in height to the corresponding ordi¬ 
nates over the springing line in the elevation, and draw 
the curves through the points so found. The groove 
for the parting bead can be marked by running a 
% in. piece around the inside of the sash-head, this 
bead being generally put in parallel. It is sometimes 
omitted altogether, the side beads being carried up 
until they die off on the head. 








PRACTICAL SOLUTIONS 


173 


To Find the Mould for the Cot Bar. Divide its cen¬ 
tre line in the elevation into equal parts, as 0, P, Q, 
R, T. Drop projectors from these into the plan, cut¬ 
ting the chord line 1 1 in o, p, q, r, t. These lines should 
be on the plan of the top sash, but are produced across 
the lower to avoid confusion with the projectors from 
the other bars. Set out the stretch out of the cot bar 




AT 

"EA-JA 

M 

R' Q' P ( 

)' P' Q' R' T 


Fig. 38. 


on the line M' 0' T', Fig. 38, and erect perpendiculars 
at the points of division, and make them equal in 
length to the correspondingly marked lines in the plan. 
The cot bar is cut out in one piece long enough to form 
the two upright sides as well as the arch. The straight 
parts are worked nearly to the springing, and the bar 
wdiich is got out wider in the centre is then steamed 
and bent around a drum, and afterwards cut to the 
mould (Fig. 38) and then rebated and moulded. The 
arched bar should not be mortised for the radial bars, 
but the latter scribed over it and screwed through from 
inside. 

To Find the Mould for the Radial Bars. Divide the 
centre line of the bar into equal parts, as 1, 2, 3, 4, 5, 
Fig. 30, and project the points into the plan, cutting 
the chord line J J in 1', 2', 3', 4', 5'. Erect perpendicu¬ 
lars upon the centre line from the points of division, 
and make them equal in length to the distance of the 
corresponding points in the plan from the chord line, 
and draw the curve through the points so obtained. 














174 


MODERN CARPENTRY 


The other bar is treated in the same manner, the 
projectors being marked with full lines in the plan. 

Fig. 39. 



ta a -TTtrm-Bi . .-- .1 - ■ ■ . I 


Fig. 40. 

The soft moulds for the head linings are obtained in 
similar manner to those for the head mould, the width 
being gauged from the head itself. 
























































































PRACTICAL SOLUTIONS 


175 


A Frame Splayed Lining with Circular Soffit as 

shown in part elevation in Fig. 39, plan Fig. 40, and 
section Fig. 41. The soffit stiles are worked in the solid 



Fig. 41. 


in two pieces joined at the crown and springings. The 
rails are worked with parallel edges, their centre lines 
radiating from the centre of the elevation. Edge 




















































































1,76 


MODERN CARPENTRY 


moulds only are required for the stiles, and a developed 
mould for the panels. The method of obtaining these 
has been shown on a separate diagram, Fig. 42. 



To Obtain the Face Moulds for the Soffit Stiles. 

Draw the line E E, Fig. 42, and produce the faces of 













PRACTICAL SOLUTIONS 


177 


the jambs until they meet in point C. From C draw 
the line C D perpendicularly to E E. This line will 
contain the centres of the various moulds, which are lo¬ 
cated by producing the plans of the edges of the stile 
across it e, 1, 2, 3 being the respective centres, and the 
inside and outside faces of the jambs affording the 
necessary radii for describing the arcs A a and B b. 

To Apply the Moulds. Prepare the stuff equal in 
thickness to the distance between the lines e 1 and 2 3, 
Fig. 42. Apply the mould A to the face of the pieces 
intended for the front stile, and cut the ends to the 
mould, and square from the face. Set a bevel as at F, 
Fig. 42, and apply it on the squared ends, working 
from the lines on the face, and apply the mould a at 
the back, keeping its ends coincident with the joints, 
and at the points where the bevel lines intersect the 
face. The piece can then be cut and worked to these 
lines, and the inside edge squared from the face. The 
outside edge is at the correct bevel, and only requires 
squaring slightly on the back to form a seat for the 
grounds. The inside stile is marked and prepared sim¬ 
ilarly. 

The Development of the Conical Surface of the Soffit 

is shown on the right hand of the diagram, Fig. 42, and 
is given to explain the method of obtaining the shape 
of the veneer, but it is not actually required in the 
present construction, as the panel being necessarily 
constructed on a cylinder, its true shape is defined 
thereon, and its size is readily obtained by marking 
direct from the soffit framing when the latter is put 
together. Let the semi-circle E D E, Fig. 42, represent 
a base of the semi-cone, and the triangle E C E its ver- 


178 


MODERN CARPENTRY 


tical section. From the apex C, with the length of 
one of its sides as radius, described the arc E F, which 
make equal in length to the semi-circle E D E by step¬ 
ping lengths as previously described. Join F to C, and 
E CF is the covering of the semi-cone. The shape of 
the frustum, or portion cut off by the section line 
of the linings, is found by projecting the inner edge of 
the lining upon the side C E, and drawing the con¬ 



centric arc I J; then E I J E represents the covering 
of the frustum. Any portion of this, the panels for 
instance, is found in the same way. To draw the rails, 
divide their centre lines equally on the perimeter, and 
draw lines from the divisions to the centre as II C; 
make the edges parallel with these lines-. 

A Window With a Splayed Soffit and Splayed Jambs 
is shown in part elevation, Fig. 43, plan Fig. 44; and 








































PRACTICAL SOLUTIONS 


179 


section Fig. 45. The linings are grooved and tongned 
together, as shown in the enlarged section, Fig. 46. To 
obtain the correct bevel required for the shoulder of 
the jamb and the groove in the soffit, the lining must 



be revolved upon one of its edges until it is parallel 
with the front, when its real shape can be seen. This 
operation is shown in the diagram on the right-hand 



Fig. 45. 


half of the plan and elevation. Draw the line a b, rep¬ 
resenting the face of the jamb in plan, at the desired 
angle. Project the edges into the elevation, and in¬ 
tersect them b}^ lines c d, projected from the top and 
bottom edges of the soffit in section. This will give the 
projection of the linings in elevation. Then from point 



































iso 


MODERN CARPENTRY 


a as centre, and the width of the lining a b as radius, 
describe the arc b b', bringing the edge b into the 
same plane as the edge a. Project point b' into the 
elevation, cutting the top edge of the soffit produced in 
c'. Draw a line from c' to d, and the contained angle 
is the bevel for the top of the jamb. When the soffit 
is splayed at the same angle as the jambs, the same 
bevel will answer for both; but when the angle is 
different, as shown by the dotted lines at e, Fig. 45 , 
then the soffit also must be turned into the vertical 
plane, as shown at e' and a line drawn from that point 
to intersect the projection of the front edge of the 
jamb in F; join this point to the intersection of the 
lower edges, and the contained angle is the bevel for 
the grooves in the soffit. (See Fig. 46.) 



The Enlarging and Diminishing of Mouldings. The 

design of a moulding can be readily enlarged to any 
desired dimensions by drawing parallel lines from its 
members, and laying a strip of paper or a straight-edge 
of the required dimensions in an inclined direction be¬ 
tween the boundary lines of the top and bottom edges; 













PRACTICAL SOLUTIONS 


181 


and at the points where the straight-edge crosses the 
various lines, make marks thereon which will be points 
in the new projection, each member being increased 
proportionately to the whole. Projections drawn at 
right angles to the former from the same points will 
give data for increasing the width in like manner. 

To Diminish a Moulding. The method to be ex¬ 
plained, which is equally applicable to enlargement, is 
based upon one of the properties of a triangle, viz., if 
one of the sides of a triangle is divided into any 
number of parts, and lines drawn from the divisions 
to the opposite side of the triangle, any line parallel 
with the divided side will be divided in corresponding 
ratio. See Fig. 47, where A B C is an equilateral tri¬ 
angle, the side A B being divided into six equal parts, 
and lines drawn from these to the apex C. The two 
lines DE and FH, parallel to A B, are divided into 
the same number of parts, and each of these parts bears 
the same ratio to the whole line that the corresponding 
part bears to A B, viz., one-sixth; the application of 
this principle will now be shown. Let it be required to 
reduce the cornice shown in Fig. 48 to a similar one 
of smaller proportions. Draw parallel projectors from 
the various members to the back line A B, and upon 
this line describe an equilateral triangle. Draw lines 
from the points on the base to the apex, then set off 
upon one of the sides of the triangle from C a length 
equal to the desired height of the new cornice as at 
G or H, and from this point draw a line parallel to the 
base line. At the points where this line intersects the 
inclined division lines, draw horizontal projectors cor¬ 
responding to the originals. To obtain their length 


182 


MODERN CARPENTRY 


or amount of projection, draw the horizontal line b E, 
Fig. 48, at the level of the lowest member of the cor¬ 
nice, and upon this line drop projectors at right angles 
to it from the various members. Describe the equilat¬ 
eral triangle b i E upon this side, and draw lines from 
the divisions to the apex i. To ascertain the length 



that shall bear the same proportion to b E that the line 
G H bears to A B, place the length of b E on the line 
BA from B to F, and draw a line from F to C: the 
portion of the line G H cut off from J to II is the pro¬ 
portionate length required. Set this length off parallel 
to b E within the triangle, as before described, and 
also draw the horizontal line L II, Fig. 49, making it 


































































PRACTICAL SOLUTIONS 


183 


equal in length to a a, Fig.. 48. Upon this line set off 
the divisions as they occur on a a, noting that their 
direction is reversed in the two figures. Erect per¬ 
pendiculars from these points to intersect the previ¬ 
ously drawn horizontals, and through the intersections 
trace the new profile. The frieze and architrave are 
reduced in like manner, M B, Fig. 48, representing the 
height of the original architrave, and N II the reduc¬ 
tion. The cornice can be enlarged similarly by pro- 



Fig. 50. 


during the inclined sides of the triangle, as shown by 
the dotted lines on Fig. 48, sufficiently to enable the 
required depth to be drawn within it parallel to A B. 
One member has been enlarged to indicate the method, 
which should be clear without further explanation. 

Raking Mouldings. Fig. 50 shows the method of 
finding the true section of an inclined moulding that 
is required to mitre with a similar horizontal moulding 











184 


MODERN CARPENTRY 


at its lower end, as in pediments of doors and windows. 
The horizontal section being the more readily seen, is 
usually decided first. Let the profile in Fig. 50 repre¬ 
sent this. Divide the outline into any number of parts, 
and erect perpendiculars therefrom, to cut a horizontal 
line drawn from the intersection of the back of the 
moulding with the top edge, as at point 7. With this 
point as a centre and the vertical projectors from 1 to 7 
as radii, describe arcs cutting the top of the inclined 
mould, as shown. From these points draw perpendicu- 



Fig. 51. 


lars to the rake, to meet lines parallel to the edges of 
the inclined moulding drawn from the corresponding 
points of division in the profile, and their intersections 
will give points in the curve through which to draw 
the section of the raking mould. When the pediment 
is broken and a level moulding returned at the top, its 
section is found in a similar manner, as will be clear 
by inspection of the drawing. If the section of the 
raking moulding is given, that of the horizontal mould 
can be found by reversing the process described above. 
Fig. 51 shows the application of the method in finding 
the section of a return bead when one side is level and 






PRACTICAL SOLUTIONS 


185 


the other inclined, as on the edge of the curb of a sky¬ 
light with vertical ends. 

Sprung Mouldings. Mouldings curved in either ele¬ 
vation or plan are called “sprung,” and when these 
are used in a pediment, require the section to be deter¬ 
mined as in a raking moulding. The operation is sim¬ 
ilar to that described above up to the point where the 
back of the section is drawn perpendicular to the in¬ 
clination, but in the present case this line E x, Fig. 52, 

c 


Pig. 52. 

is drawn radiating from the centre of the curve, and 
the projectors are drawn parallel to this line. The 
parallel projectors, 1, 2, 3, 4, 5, are also described from 
the centre until they reach the line E x, when per¬ 
pendiculars to this line are raised from the points of 
intersection to meet the perpendicular projectors. 

Mitreing Straight and Curved Mouldings Together. 
If a straight mitre is required, draw the plan of the 
mouldings, as in Fig. 53, and the section of the straight 
mould at right angles to its plan as at A. Divide its 






186 


MODERN CARPENTRY 


profile into any number of parts, and from them draw 
parallels to the edges intersecting the mitre line. From 
these intersections describe arcs concentric with the 
plan of the curved mould, and at any convenient point 
thereon draw a line radial from the centre. Erect 
perpendiculars on this line from the points where the 
arcs intersect it, and make them equal in height to the 
corresponding lines on the section of the straight 
moulding A, and these will be points in the profile of 



Fig. 53. 


the curved moulding B. When it is required that the 
section of both mouldings shall be alike, a circular 
mitre is necessary, and its true shape is obtained as 
shown in Fig. 53. Draw the plan and divide the profile 
of the straight moulding as before, drawing parallels 
to the edge towards the seat of the mitre. Upon a line 
drawn through the centre of the curved moulding set 
off divisions equal and similar to those on the straight 
part, as 1 to 8 in the drawing. From the centre of the 






















PRACTICAL SOLUTIONS 


187 


curve describe arcs passing through these points, and 
through the points of intersection of these arcs with 
the parallel projectors, draw a curve which will be the 
true shape of the mitre. Cut a saddle templet to this 
shape, and use it to mark the mouldings and guide the 
chisel in cutting. 



To Obtain the Section of a Sash Bar Raking in Plan. 

Fig. 54 represents the plan of a shop front sash with 
bars in the angles. On the left hand is shown the sec- 






























188 


MODERN CARPENTRY 


tion of the stile or rail into which the bars have to 
mitre. Divide the profile of moulding into a number 
of parts, as shown from 1 to 6, and from these points 
draw parallels to the sides of the rails intersecting the 
centre line of the bar. Also draw perpendiculars from 
the same ponts to any line at right angles to them, as 
at A. Draw a line at right angles to the centre line of 
the bar, and on it set off the divisions from 1 to 6 as 
at A. Draw projectors from these points parallel to 
the centre line of the bar, and where they intersect 
the correspondingly numbered lines drawn parallel 
with the sides of the sash will be points in the curve of 
the section of the bar. It will be noticed that there is 
no fillet or square shown on the bar, and that in trans¬ 
ferring the points from the line A they must be re¬ 
versed on each side of the centre. Should a fillet be 
required on the bar, additional thickness must be given 
for the purpose. Three methods of forming the rebates 
in the bar are shown, the screwed saddle beads being 
the best for securing the glass. 

Interior Shutters Include Folding, Sliding, Balanced, 
and Rolling. Exterior Consist of Hanging, Lifting, 
Spring, and Venetian Shutters. 

FOLDING, or as they are frequently called, BOX¬ 
ING SHUTTERS, because they fold into a boxing or 
recess formed between the window frames and the 
walls, are composed of a number of narrow leaves, 
framed or plain, as their size may determine, rebated 
and hinged to each other and to the window frames. 
r I hey should be of such size that when opened out they 
will cover the entire light space of the sash frame and 
a margin of a % in. in addition. Care must be taken 


PRACTICAL SOLUTIONS 


189 


to make them parallel, or they will not swing clear 
at the ends; and as a further precaution, they should 
not be carried right from soffit to window board, but 
have clearance pieces interposed at their ends about 
% in. thick. The outer leaf, which is always framed, 
is termed a shutter • the others are termed flaps. It is 
not advisable to make the shutters less than 1*4 in. 
thick, and flaps over 8 in. wide should be framed; those 
less than 8 in. may be solid, but should be mitre 
clamped to prevent warping. In a superior class of 
work the boxings are provided with cover flaps which 
. conceal the shutters when folded, and fill the void when 
they are opened out. The sizes and arrangements of 
the framing are determined by the general finishings 
of the apartment, but it is usual to make the stiles ol 5 
the front shutters range with those of the soffit and 
elbow linings. When Venetian or other blinds are used 
inside, provision is made for them by constructing a 
block frame from 2to 3 in. thick inside the window 
frame and the shutters are hung to this. 

The leaves are hung to each other with wide hinges 
called back flaps, that screw on the face of the leaves, 
there not being sufficient surface on the edges for butt 
hinges. In setting out the depth of boxings, at least 
V 8 in. should be allowed between each shutter to pro¬ 
vide room for the fittings; the shutters are fastened 
by a flat iron bar hung on a pivot plate fixed on the 
inner left-hand leaf, and having a projecting stud at 
its other end which fits into a slotted plate, and it is 
kept in this position by a cam or button. Long shutters 
are made in two lengths, the joint coming opposite the 
meeting rails of the sashes; these are sometimes re¬ 
bated together at the ends. 


190 


MODERN CARPENTRY 



Fig. 55. 






















































































































































































































PRACTICAL SOLUTIONS 


191 


Fig. 55 is a sectional elevation of a Window Fitted 
with Boxing Shutters having a cover flap and spaces 
for a blind and a curtain. One-half of the elevation 
shows the shutters opened out and the front of the 
finishings removed, showing the construction of the 
boxings, &c. The plan, Fig. 56, is divided similarly, 
one-half showing the shutters folded back, with por- 



Fig. 56. 


tions of soffit cornice, &c.; the other half gives the plan 
and sections of the lower parts, the dotted lines show¬ 
ing the window back. 

Fig. 57 is a vertical section and Fig. 58 an enlarged 
section through the boxings. The framed pilaster cov¬ 
ering the boxing is cut at the level of the window 
board, and hung to the stud A', this being necessary 
for the cover to clear the shutters when open (see 
dotted lines on opposite half). The cover flap closes 
into rebates at the top and bottom, as shown in sec- 
























































































192 MODERN CARPENTRY 



tion in Fig. 55. The window 
back is carried behind the 
elbows, and grooved to receive 
the latter. The rails of the 
soffit must be wide enough to 
cover the boxings, and should 
have the boxing back tongued 



into it, as shown in the section, 
Fig. 55. When the linings of 
an opening run from soffit to 
the floor uninterruptedly, they 
are called jamb linings; but 
when they commence, as in 
the present instance, under the 


































































































































































PRACTICAL SOLUTIONS 


193 


window board, they are termed elbow linings, the cor¬ 
responding framing under the window being the win¬ 
dow back. 

Sliding Shutters are used instead of folding shutters 
in thin walls, and consist of thin panelled frames run¬ 
ning between guides or rails fixed on the soffit and 
window board, which are made wider than usual for 



Fig. 59. 


Fig. 60. 


that purpose. When open, they lie upon the face of 
the wall adjacent to the opening. They are so seldom 
used that they do not call for illustration. " 

Balanced or Lifting Shutters are shown in section in 
Fig. 61, and plan in Fig. 62. They consist of thin pan¬ 
elled frames, the full width of and each half the height 
























































194 


MODERN CARPENTRY 



Fig. 61. 


of the window, hung with weights in a cased frame in 
a similar manner to a pair of sashes. The frame ex¬ 
tends the whole height of the window, and is carried 
down behind the floor joists to the set off. The win- 



























































































PRACTICAL SOLUTIONS 


195 

































































































































196 


MODERN CARPENTRY 


dow board is hinged to the front panelling or shutter 
back, so that when the former is lifted, the shutters 
can pass down behind the back out of sight. Cover 
flaps hung to the outer linings on each side close over 
the face of the pulley stiles and hide the cords. When 
it is desired to close the shutters, the cover flaps are 
opened out flat, as shown to the left of the plan. The 


Steel bam 



Fig. 63. 


window board can then be lifted, and the shutters run 
up, a pair of flush rings being inserted in the top edge 
of each for that purpose. The window board is next 
shut down, the inside shutter brought down upon it, 
the outside one pushed tight up to the head, and the 
meeting rails, which overlap an inch, fastened with a 
thumb-screw. The bottom rail of the upper shutter 
is made an inch wider than the other, for the pur¬ 
pose of showing an equal margin when overlapped. 
Square lead weights have generally to be used for 


















PRACTICAL SOLUTIONS 


197 


these shutters in consequence of their comparative 
heaviness. 

Rolling 1 or Spring Shutters are made in iron, steel 
and painted or polished woods, and though somewhat 
monotonous in appearance, are, in consequence of 
their convenience in opening and closing, and in the 
case of the metal kinds, the additional security against 
fire and burglary, fast superseding all other kinds, 
both for internal and external use, especially in shops 
and public buildings. They consist, in the case of the 
wood varieties, of a series of laths of plano-convex, 
double convex, oval, or ogee section, fastened together 
by thin steel or copper bands passing through mortises 
in the centre of the thickness, as shown in Fig. 63, 
and secured at their upper ends to the spring con¬ 
tainer. These metal bands are supplemented by sev¬ 
eral waterproof bands of flax webbing glued to the 
back sides of the laths. The upper edge of each lath 
is rounded, and fits into a corresponding hollow in 
the lower edge of the one above it; this peculiar over¬ 
lapping joint, whilst preventing the passage of light, 
etc!, between the laths, readily yields when 
the shutter is coiled around the barrel. The spring 
barrel is usually made of tinned iron plate, and en¬ 
cases a stout spiral spring wound round an iron man¬ 
drel with squared ends to which a key is fitted for 
winding the spring up. The ends of the mandrel pro¬ 
ject from the case and are fixed in bracket plates at 
each end of the shutter—in the case of shop fronts, 
in a box or recess behind the fascia; one end of the 
spring is secured to the barrel, the other end to the 
mandrel; the shutter is secured to the barrel by the 


198 


MODERN CARPENTRY 



Fig. 64. 





















































PRACTICAL SOLUTIONS 


199 


metal bands mentioned above, and the normal con¬ 
dition of things is that when the shutter is coiled up 
the spring is unwound. The pulling down of the 
shutter winds up the spring, and the tension is so 
arranged that it does not quite overcome the weight 
of the shutter and the friction when the shutter is 
down, so that the latter must be assisted up with a 
long arm. Great care should be taken in fixing these 
shutters to arrange the barrel perfectly level and par¬ 
allel with the front, and to securely fix the brackets. 
These may be bolted to the girder or breastsummer, 
or screwed to the fixings plugged in the wall, as shown 
by dotted lines in Fig. 64. Where possible a wood 
groove should be formed on each side of the openings 
for the shutters to work in, but iron channels 
(Fig. 65) are frequently used. These are 
cemented into a chase in the wall or pilaster. 

Fig. 64 is a section through a shop fascia, 
showing shutter and blind barrels fixed to 
the face of the wall, where no provision 
has been made beneath the girder. Fig. 66 shows the 
adaption of one of these shutters to the inside of a win¬ 
dow; the barrel is fixed in a seat formed beneath the 
window, the top of which is hinged, to gain access to 
the coil. The minimum space required for a coil, for 
a shutter about 6 ft. high is 1014 in. A friction roller 
F should be fixed close to the back rail to prevent 
chafing as the coil unwinds; the shutter is lifted by 
means of a flush ring in the L iron bar at the top, and 
this ring engages with a tilting hook in the soffit to 
keep the shutter up. The metal varieties of these shut¬ 
ters are wound up by the aid of balance weights or 
bevel wheel gearing. 









200 


MODERN CARPENTRY 
























































PRACTICAL SOLUTIONS 


201 


WINDOWS GENERALLY. 

Size and Position. The sizes and positions of win¬ 
dow openings are influenced by the size of the rooms, 
and the purposes for which the building is used. For 
the sake of ventilation, and also to secure good light¬ 
ing, the windows should be placed at as great a height 
as the construction of the room will allow. In dwell¬ 
ing-houses the height of the sill is usually about 2' 6" 
above the inside floor level. 

Construction. The framework holding the glass of 
the window may be fixed or movable. It must be so 
prepared that the glass can be replaced easily when 
necessary. In warehouses, workshops and similar 
buildings, the frames holding the glass are often fixed 
as Fast Sheets (Fig. 66^). As, however, this arrange¬ 
ment affords no means of ventilation, it is more usual to 
have the glass fixed in lighter frames called Sashes. If 
the sashes are hung to solid rebated frames, and open 
as doors do, the windows are called Casement Sashes. 
If they slide vertically and are balanced by weights or 
by each other, the window is a Sash and Frame Win¬ 
dows. Other methods of arranging sashes, either 
hinged, pivoted or made to slide past each other, are 
described in detail later. 

Sashes. The terms used for the various parts of 
sashes and fast sheets are somewhat similar to those 
employed in describing doors. Thus, the Styles are the 
outer uprights, and the Rails are the main horizontal 
cross-pieces: top rails, meeting rails, and bottom rails 
being distinguished. Any intermediate members, 
whether vertical or horizontal, are named Bars. 


202 


MODERN CARPENTRY 



Rg. 66'A 


Sashes are from iy 2 to 3 inches thick. The inner edge 
of the outer face is Rebated to receive the glass. The 
inner face is left either square, chamfered, or moulded; 



Fig. 67. Fig. 68. Fig. 69. Fig. 70. 

two common forms of moulding are lamb’s-tongue (Fig. 
68) and ovolo (Fig. 69). The size of the rebate is indi- 
























































PRACTICAL SOLUTIONS 


203 


cated in Pig. 70; it varies with the thickness of the sash, 
its depth being always a little more than one-third this 
thickness. The width of the rebate varies from a quar¬ 
ter of an inch to half an inch, and the mould is usually 
sunk the same depth as the rebate. This last fact is of 
some importance, as it affects the shoulder lines; and 
with hand work it influences the amount of labor in 
the making of the sashes. 

As little material as possible is used in the sashes, in 
order that the light shall not be interfered with. In gen¬ 
eral, the styles and top rail are square in section before 
being rebfted and moulded. In casement sashes, how¬ 
ever, it is often advisable to have the outer styles a 
little wider than the thickness, especially when they are 
tongued into the frame. The width of the bottom rail is 
from one and a half to twice the thickness of the sash. 
Sash bars, which require rebating and moulding on both 
sides, should be as narrow as possible, in order not to 
interrupt the light. They are usually from five-eighths 
of an inch to one and a quarter inches wide. 

Joints of Sashes. The sashes are framed together by 
means of the Mortise and Tenon Joint (Fig. 71). The 
proportions of the thickness and width of tenons, 
haunched tenons, &c., are to a large extent applicable 
here. Hardwood cross-tongues are sometimes inserted 
to strengthen the joints while thick sashes should have 
Double Tenons. An alternative to halving in sash bars 
is to arrange that the bar which is subjected to the 
greater stress—as for example, the vertical bars in slid¬ 
ing sashes, and the horizontal bars in hinged casement 
sashes—shall be continuous; this continuous bar is mor¬ 
tised to receive the other, which is scribed i, e, cut to 


204 


MODERN CARPENTRY 


fit the first, and on which the short tenons are left. This 
method is called Franking the Sash Bars, and is illus¬ 
trated in Fig. 72. 




Fig. 72. 


Casement Windows. Casement windows may be 
hinged in such a manner that they open either inwards 
or outwards. They may consist either of one sash, or of 
folding sashes, and are hung with butt hinges to solid 
rebated frames. These Frames consist of jambs, head 









































PRACTICAL SOLUTIONS 


205 


and sill. The head and sill “run through” and are 

mortised near the ends to receive tenons formed on the 

ends of the jambs. The upper surface of the sill is 

weathered to throw off rain water. Casement windows 

which reach to the floor are usuallv called French Case- 

«/ 

ments. Their sashes require an extra depth of bottom 
rail. 

Casement Sashes Opening Inwards. Figs. 73, 74 show 
the elevation and vertical and horizontal sections, of 
a window opening in a 14" brick wall fitted with a 
casement window having folding sashes to open in¬ 
wards. In this class of window the frame is rebated for 
the sashes on the inner side. Each sash has, on the 
outer edge of the outer style, a semi-circular tongue, 
which fits into a corresponding groove in the jamb of 
the frame. This tongue renders the vertical joint be¬ 
tween the sash and frame more likely to be weather 
proof; it is to provide for the tongue that the extra 
width of style already referred to is necessary. The 
tongue, however, is often omitted, as in Fig. 77. It 
will be seen readily that, if the sash were in one width, 
it would be impossible to have a tongue on more than 
one edge of it. With casement sashes opening inwards 
the greatest difficulty is found, however, in making a 
water-tight joint between the bottom rail of the sash 
and the sill of the frame. Figs. 77 and 78 show two 
methods by which this may be accomplished. An es¬ 
sential feature of all these sashes is a small groove or 
Throating on the under edge of the bottom rail; this 
prevents the water from getting through. The groove 
in the rebate of the sill (Fig. 78) is provided to collect 
any water that may drive through the joint. This 


206 


MODERN CARPENTRY 



Fig. 73. 


Fig. 74. 




K N^ X v \ O v N\'y' f / 


Enlarged Verfical 
Section. 

Fig. 75. 



Fig. 76. 






































































































































































PRACTICAL SOLUTIONS 207 


water escapes through the hole bored in the centre of 
the sill. 



When easement sashes are hung after the manner of 
folding doors, the vertical joint between the meeting 
styles is rebated. Alternative methods of rebating 


Sash opens 
inwards 



Fig. 78. 



Fig. 79. 


are shown in Fig. 79 and 80. Fig. 80 is known as a 
Hook Joint and is the better one. 







































208 


MODERN CARPENTRY 


Casement Sashes Opening Outwards. These are more 
easily made weatherproof than inward-opening sashes. 
The chief objections to their adoption are that they are 
not easily accessible for cleaning the outside, especial- 
1 y in upper rooms .and that they are also liable, when 
left open, to be damaged by high winds and to let in 
the rain during a storm. Fig. 81 is a sketch of one 
corner of such a window. It will be noticed that these 



frames, like door liames, have the exposes arrises 
moulded in various ways, and that the sashes may 
either be hung flush with one face of the frame (as in 
Figs. 76 and 77), or fit in the thickness of the frame 
(Figs. 78 and 81). The sill shows to be Double Sunk, 
i. e., to have the upper surface—upon which the bot¬ 
tom rail of the sash fits—rebated with two slopes 
(weatherings). 



































PRACTICAL SOLUTIONS 


209 


Sash and Frame Window. In this class of window, 
which is by far the most common, because it is easily 
made weather proof, there are Two Sashes, which slide 
past each other in vertical grooves, and are usually 



balanced by iron or leaden weights. As will be seen 
from Fig. 82 the frames form cases or boxes in which 
the weights are suspended. They are hence called Cased 
Frames. Pulley Styles (Fig. 86) take the place of the 
solid rebated jambs of casement windows. The pulley 
styles, Outside and Inside Linings, and Back Lining 
(Fig. 82) together form a box which is subdivided by a 
vertical Parting Slip suspended as shown in Fig. 82. In 
superior window frames of this kind, the pulley styles 
and linings are tongued and grooved together as shown 
in Fig. 83. In commoner work the tongues and grooves 
are often omitted. The frame must be so constructed 
























































































210 


MODERN CARPENTRY 


that the sashes can be removed easily for the purpose 
of replacing broken sash-lines. To enable this to be 
done, the edge of the inside lining is either made flush 
with the face of the pulley style (Fig. 83), or it is re¬ 
bated slightly as shown in Fig. 83. The edge of the out¬ 
side lining projects for a distance of about three-quar¬ 
ters of an inch beyond the face of the pulley style, to 
form a rebate against which the outer (upper) sash 
slides. The outer sash is kept in position by the Part- 



thick 
Staffbead /i>>% * 



Fig. 83. 


Fig. 83 y. 


ing Lath (Fig. 82") which fits into a groove in the pul¬ 
ley style. The groove for the inner (lower) sash is 
formed by the parting lath and a Staff Bead or Stop 
Bead which is secured by screws. The staff bead on 
the sill is often made from two to three inches deep to 
allow the lower sash to be raised sufficiently for ven¬ 
tilation at the meeting rails without causing a draught 
at thebottom (Fig. 83). 

A vertical section through the head of the frame is 
similar to a horizontal section across the pulley style, 

















































PRACTICAL SOLUTIONS 211 

except that the back lining and parting slip are of 
course absent. 

The sill of the frame is solid and weathered, and 
should always be of hardwood, preferably oak or teak. 
T he sill has a width equal to the full thickness of the 
frame. When the weathering has two steppings, it is 
known as a Double Sunk Sill. An alternative to the 
plan of having the width of the sill of the full thickness 
of the frame, is to arrange it so that the outside edge 
is flush with the outside face of the bottom sash, as 
shown in Fig. 89. With a sill arranged in this manner, 
and double sunk, there is less danger of water driving 
through the joint between the sash and the sill than 
with a sill the full thickness of the frame. In order 




Pulley style, 


P',9 C ^ 




Fig. 84. 


to render watertight the joint between the wooden 
and stone sills of window frames, a metal tongue is 
often fixed into corresponding grooves cut into the un¬ 
der side of the wooden sill and the upper surface of 
the stone sill. A rebated joint between the two sills 
serves the same purpose as the metal tongue. 

Fig. 84 shows the methods of fixing the pulley style 
into the head and sill respectively, when the width of 
the sill is equal to the full thickness of the frame. The 
Pulleys on which the sash lines run—sash or axle pul¬ 
leys are fixed in mortises near the upper ends of the 











212 


MODERN CARPENTRY 


pulley styles. It is also necessary to have a removable 
piece in the lower part of each pulley style, to allow 
of access to the weights. This piece is named the 
Pocket Piece. It may be cut as shown in Fig. 85; its 
position is then behind the lower sash, and it is hidden 



from view when the window is closed. Or, the pocket 
piece may be in the middle of the pulley style as shown 
in Fig. 84; the vertical joints between the pocket piece 
and the pulley style are then V shaped to prevent dam¬ 
age to the paint in case of removal. 



Sashes. The only difference between the joints of 
sliding sashes and those of the casement sashes already 
described is in the construction of the meeting rails. 
Each of the meeting rails is made thicker than the sash 
to the extent of the thickness of the parting lath; other¬ 
wise there would be a space between them equal to the 




























































PRACTICAL SOLUTIONS 


213 


thickness of the parting lath. The joint between them 
may be rebated or splayed. The angle joints between 
the ends of the sash styles and the meeting rails are 




Style. 
Fig. 86. 


often dovetailed as shown in Fig. 86. They are, how¬ 
ever, stronger if the styles are made a little longer, 
the projecting part being moulded, and mortise and 

























'Architrave 


214 


MODERN CARPENTRY 


tenon joints used as shown in Figs. 86 and 87. The 
projecting ends of the styles are called Joggles; they 
assist in enabling the sashes, especially in wide win¬ 
dows, to slide more freely. When as is usually the case, 
both sashes slide and are balanced by weights, the 



Fig. 87. 

window is known as a Double-Hung sash and frame 
window. If one sash only slides, and the other is fixed 
in the frame, the window is Single-Hung. Figs. 78 to 
80 show the details of a sash and frame window fixed 
in a one-and-a-lialf-brick-thick wall and having a stone 
head and sill. 






















































































PRACTICAL SOLUTIONS 


215 


For the sake of appearance, or when it is required to 
have wider windows than can be arranged with one 
pair of sashes, two or three pairs of sashes are often 
constructed side by side in the same frame. When 
three pairs of sashes are used, it is usual to have the 
middle pair wider than the others; such a combination 
(Fig. 73) is named a Venetian Window. The vertical 
divisions between adjacent pairs of sashes are called 
Mullions. These mullions may be constructed in sev¬ 
eral different ways. If the middle pair of sashes only is 
required to slide, the mullions may be solid from l 1 /^" 
to 2" thick, and the .sash-cord conducted by means of 
additional pulleys to the boxes, which are at the outer 
edges of the frame. Figs. 84 and 86 show this arrange¬ 
ment. If it is desirable to have all the sashes to slide, 
the mullions must be hollow to provide room for the 
weights. Figs. 85 and 87 show details of a mullion with 
provision made for one weight to balance the two sashes 
adjacent to it. With this arrangement the sash-cord 
passes round a pulley fixed into the upper end of the 
weight. If stone mullions are used in the window open¬ 
ing, separate boxings may be made so that each pair of 
sashes is hung independently as shown in Fig. 83, and 
the window becomes, as it were, two or three—as the 
case may be—separate window frames, with the sill 
and head each in one length for the sake of strength. 

The Hanging of Vertical Sliding Sashes. As shown 
in illustrations already given, the sashes of sash and 
frame windows are balanced by cast-iron or leaden 
weights. The best cord is employed for hanging sashes 
of ordinary size, while for very heavy sashes the sash 
lines are often of steel or copper. The staff bead and 


216 


MODERN CARPENTRY 


parting bead having been removed, the cords are passed 
over the axle pulleys (which are best of brass to pre¬ 
vent corrosion) and are tied to the upper ends of the 
weights. The weights are passed through the pocket 
holes and suspended in the boxes. The pocket pieces 
having been replaced, the upper sash, which slides in 
the outer groove, is hung first, the free ends of the 
cords being either nailed into grooves in the outer 
edges of the sash or secured by knotting the ends after 
passing them through holes bored into the styles of the 
sash. r lhe upper sash having been hung, the parting 
laths are fixed into the grooves in the pulley styles, and 
the lower (inner) sash is hung in a similar manner, 
after which the staff beads are screwed in position. 
Care should be taken to have the cords of the right 
lengths; if the cords for the upper sash are too long 
the weights will touch the bottom of the frame, and 
cease to balance the weight of the sash before the lat¬ 
ter is closed. If the cords for the lower sash are too 
short, the weights will come in contact with the axle 
pulleys, and thus prevent it from closing. Several dif¬ 
ferent devices for hanging sashes—the objects of 
which are either to render unnecessary the use of 
weights or to facilitate the cleaning of the outside of 
the window have been patented, and are in more or 
less general use. A detailed description of these is, 
however, beyond the scope of this book. 

Bay Windows. A bay window is one that projects 
beyond the face of the wall. The side lights may be 
either splayed or at right angles to the front. The 
window openings may be formed by having stone or 
brick mull ions or piers at the angles, against which 


PRACTICAL SOLUTIONS 


217 


the window frames are fixed, or the wooden framework 
of the window may be complete in itself. When the 
latter is the case, it is usual to have stone or brick 
work to the sill level as shown in Fig. 88. Bay win¬ 



dows naturally lend themselves to decorative treat¬ 
ment. With the addition of masonry or brickwork they 
often assume a massive and bold appearance. When 
usually by a wooden cornice, and the wooden roof is 











































































































































































218 


MODERN CARPENTRY 


covered with lead, slates or tiles. The window frames 
may he arranged as fixed lights, sash and frame, or 



easements. The most usual arrangement is to have 
the lower lights fixed, and the upper ones as sashes 
hinged to open for ventilating purposes. Figs. 88 to 90 
show the details of a bay window with splayed side 











































































































































































































































































































PRACTICAL SOLUTIONS 


219 


lights, the upper side lights being hinged on the tran¬ 
som to open inwards. 

Windows With Curved Heads. When a window 
opening is surmounted by an arch, the top of the win¬ 
dow frame requires to be off the same curvature as the 
under side (soffit) of the arch. In the case of fixed 



Horizontal Sect tony 



Fig. 90. 


sashes, or of solid frames with casement sashes, the 
head of the frame is “cut out of the solid.” A head 
which, owing to the size of the curve, cannot easily be 
obtained in one piece, is built up of segments, the 



joints being radial to the curve, and secured by hard¬ 
wood keys. As an alternative method, the head may 
be built up of two thicknesses*—with overlapping 
joints—and secured together by screws. 












































220 


MODERN CARPENTRY 


A sash and frame window in such an opening may 
have only the outside lining cut to the curve of the 
arch, the inner side of the frame being left square. The 
upper sash will then require a top rail with a straight 
upper edge and a curved lower edge, as shown in Fig- 
91 . 



Y hen the head of the frame has to be curved, it 
may 

(1) be built up of two thicknesses with overlapping 
joints, and secured by screws; it may 

(2) be formed of three thicknesses of thin material, 
bent upon a block of the correct radius, and well glued 
and screwed together; or 

(3) the head may be of the same thickness as the 
pulley styles, with trenches cut out of the back (up¬ 
per) side, leaving only a veneer on the face-side under 
the trenches. Wooden keys are glued and driven into 
the trenches after the head has been bent upon a block 
to the required shape. 

A strip of stout canvas glued over the upper side 
will strengthen the whole materially. The outside and 
























PRACTICAL SOLUTIONS 


221 


inside linings are in such a case cut to the required 
curvature, and when nailed in position hold the head 
in shape. The end joints of the linings may have hard¬ 
wood cross-tongues. 

Shop Windows. The main object in view in the con¬ 
struction of shop windows is to admit the maximum of 
light, and to give opportunity for an effective display 
of the goods. The glass is in large sheets, and there¬ 
fore is specially thick to secure the necessary strength. 
Shop windows are usually arranged as fast sheets, with 
provision for ventilation at the top. The glass is held 
in position by wooden fillets, and is fixed from the in¬ 
ner side. The chief constructional variations are found 
in the pilasters, cornice, provision for sign-board, sun- 
blind, and the arrangement of the side windows. Figs. 
92 to 95 show the details of a typical example. 

The Fixing of Window Frames. Window frames 
may be built into the wall—which has usually a re¬ 
cessed opening to receive them—as the brickwork pro¬ 
ceeds, or they may be fixed later. In the former case, 
the ends of the sill and head project and form Horns, 
which are built into the brickwork and help to secure 
the frame. Wooden bricks or slips may also be built in¬ 
to the wall, the frames being nailed to them. 

In the latter case, the frames are secured by wooden 
Wedges, which are driven tightly between the frame 
and the wall. These wedges should be inserted only at 
the ends of the head and sill and directly above the 
jambs; otherwise the frame might be so strained as to 
interfere with the sliding of the sashes. Window 
frames as well as door frames should be bedded against 
a layer of hair-mortar placed in the recess. 


222 MODERN CARPENTRY 


Fig. 92. 



Fig. 94. 


FJg. 93. 































































































































































































































PRACTICAL SOLUTIONS 222. 



Sketch of part of a Sash and Frame Window, showing 
Panelled and Moulded Jamb-lining, etc. 


Fig - . 95. 


Linings. When window frames are not of sufficient 
thickness to come flush with the inner face of the wall, 
the plaster may be returned round the brickwork and 





































































224 


MODERN CARPENTRY 


finished against the frame, or a narrow fillet of wood 
may be scribed to the wall and nailed to the frame. In 
dwelling-houses, however, the more usual way is to 
fix linings similar to those used for outer door frames, 
idle width of the linings depends upon the thickness 
of the wall; they should project beyond the inner face 
of the wall for a distance equal to the thickness of the 



Fig. 96. 


plaster, and are usually splayed so that they will not 
interfere with the admission of light. The inside of 
window and door openings usually are finished similar¬ 
ly; thus, the Architrave, or Band Moulding, which is 
secured to the edge of the linings and to rough wooden 
Grounds is fixed along the sides and top in both cases. 






























PRACTICAL SOLUTIONS 


225 


The bottom of the window opening is finished with a 
Window Board which is tongned into the sill of the 
frame. The board is about 1% inches thick, and is 
made wide enough to project beyond the surface of the 
plaster for a distance of about 18 inches. The project¬ 
ing edge is nosed (rounded) or moulded. It is longer 
than'the opening, to allow the lower ends of the archi¬ 
trave to rest upon it. 

When the walls are thick, the linings are often 
framed and panelled. Such linings may terminate on 
a window board at the sill level, or the inner side of 
the wall may be recessed below the sill level and the 
linings carried to the floor as shown in Pig. 96. 

Window Shutters. Although not used to the same 
extent as formerlv, wooden window shutters are fitted 
occasionally to close up the window opening. Window 
shutters, which are arranged generally on the inner side 
of the window, may be hinged as box shutters, or may 
be vertically sliding shutters. 

Box Shutters consist of a number of leaves or nar¬ 
row frames which are rebated and hinged together, an 
equal number being on each side of the window open¬ 
ing, the outer ones on each side being hung to the win¬ 
dow frame. When closed they together fill the width 
of the window-space, and when open they fold behind 
each other so that the front one forms the jamb lining 
of the window frame. If the walls are thick, the shut¬ 
ters can be arranged to fold in the thickness of the 
wall; if the wall is a thin one it is necessary to con¬ 
struct projecting boxes into which the shutters fold. 
The nature of the framing of the shutters depends upon 
the surrounding work; it is usual to have the outer sur- 


226 


MODERN CARPENTRY 


face framed and moulded, and the inside finished bead- 
flush. The arrangement of box shutters requires that 
the shutters on the same side shall vary in width so 
that they will fold into the boxes on each side of the 
window, the outermost shutter (which is the widest) 
then acting as the window lining. Fig. 96 shows a hori¬ 
zontal section through one side of a window, showing 
hinged shutters folding so that a splayed lining is ob¬ 
tained. Fig. 97 shows hinged shutters consisting of one 



narrow and one wide shutter on each side of the open¬ 
ing. This arrangement is suitable for a thin wall, 
where it is undesirable to have boxes for the shutters 
projecting beyond the face of the wall. For hanging 
window shutters it is usual to use back-flap hinges; the 
joint at the corner of the shutters in Fig. 97 is named 
a rule joint. 

Sliding Shutters, working in vertical grooves and 
balanced by weights, are sometimes used. They require 
that the wall under the window sill shall be recessed; 
the floor also often needs trimming to allow space for 



































PRACTICAL SOLUTIONS 


227 


them to slide sufficiently 
low. To hide the grooves 
in which the shutters 
slide, thin vertical flaps 
are hung to the window 
frame, and the window 
board is also hinged at 
the front edge to allow 
the shutters to slide be- 



Fig. 99. 


low the sill. Figs. 98 and 
99 are sections of vertical 
sliding shutters. 

Hinged Skylights. 

Skylights which are 
hinged to open are fitted 
upon the upper edge 
of a Curb or frame 
fixed in the plane of the 
roof, the common rafters 


























































































































































































228 


MODERN CARPENTRY 


being “trimmed” to the required size to receive the curb. . 
The curb is made from material 1^4 t° 2 inches thick, 
and of width such that its upper edge stands from 4 to 
6 inches above the plane of the roof. The Angle Joints 
of the curb may be dovetailed or tongued and nailed. 
The Sash Frame rests upon the upper edge of the curb; 
it is from 2 to 2% inches thick, and consists of stiles 
and top rail of the same thickness, and a bottom rail 
which is thinner than the stiles by the depth of the 
rebate. Bars are inserted in the direction of the slope 
of the roof, and the butt hinges used for hanging the 
sash are invariably fixed as the under side of the top 
rail. 

The joints between the curb and the roofing slates or 
tiles are made weatherproof with sheet lead. At the 
upper end—the hack of the curb—a small Lead Gutter 
is formed, with the lead going underneath the slates 
and overlapping the upper edge of the curb. The sides 
of the curb may he flashed with soakers—short lengths 

V 

of sheet lead which are worked in between the slates— 
or the joint may be made with one strip of lead forming 
a small gutter down the side of the curb. In either 
case the lead overlaps the upper edge of the curb. At 
the lower end of the curb, the lead overlaps the slates. 
To prevent water from rising between the glass and the 
upper side of the bottom rail, sinkings are cut into the 
rail. (See Fig. 100). 

Dormer Windows. Instead of having the light in or 
parallel to the plane of the roof, it affords a more ar¬ 
tistic treatment of the roof, and often gives a better re¬ 
sult in lighting, if the window is fixed vertically. The 
general arrangement of the framing, as well as of the 


PRACTICAL SOLUTIONS 


229 


sashes, depends upon the kind of roof, the width of the 
window recpiired, and the general style of architecture 
of the building. 

The construction of a dormer window necessitates 
trimming of the rafters, and the arrangement of pro¬ 
jecting framework, the front of which consists of Cor- 



,Sketch of part of the Roof of a Building, showing a Hinged 
Skylight and a Dormer Window. 

Fig. 100. 


ner Posts and Crossrails —rebated to receive hinged 
sashes—which are connected to the main roof by other 
crossrails and by Braces. This framework is surmount¬ 
ed by a Roof which may be either ridged, or curved out¬ 
line, or flat. By arranging a ridged roof to overhang, 













230 


MODERN CARPENTRY 


and adding suitable Barge Boards and Finial (Pig. 101) 
a dormer window may be made to improve tlie general 
appearance of the roof of a building. The sides of the 
dormer may be either boarded and covered with the 
same kind of material as the roof, or they may be 
framed for sidelights. 


CD 



As dormer windows are generally in exposed posi¬ 
tions, and the sashes are arranged as casements to open, 
their efficiency depends largely upon the perfection of 
the joints between the sashes and the frame. It ought 
to be mentioned however, that with sashes hung fold¬ 
ing, semicircular tongues on their hanging stiles are 
by far the best. Figs. 101 and 102 give the details of 


























































































PRACTICAL SOLUTIONS 


231 


a dormer window, with sidelights, fixed in a roof of 
ordinary pitch. The sashes, which are hung folding, 
open inwards. The roof may be boarded and covered 
with lead, or it may be covered with slates or tiles. The 
joints between the roofing slates of the main roof, and 
the roof and sides of the dormer, are made weather¬ 
proof with sheet lead flashings. Figs. 103 and 104 show 
a dormer window fixed in a Mansard roof; in this ex¬ 
ample there are no side lights. 



Fig. 103. 



Large Skylights and Lantern Lights. For lighting 
the well of a large staircase, or a room which, for some 
reason, cannot be lighted with side windows, specially 
large skylights are often necessary. These are of a 
more elaborate construction than the skylights already 
described; they vary considerably in size, shape and 
design; the plan may be rectangular, polygonal, cir- 






























































232 


MODERN CARPENTRY 


cular, or elliptical, and the outline may be pyramidal, 
conical, or spherical. The framework may be of either 
wood or iron. To support such a skylight, a strong 
wooden curb is framed into the roof, and projects from 
6 to 9 inches above the roof surface. The joints be¬ 
tween the curb and the roof are made watertight with 
sheet lead. The framework of the skylight may con¬ 
sist of rebated quartering; with separate lights which 
fit into the rebates of the framing: or the sashes them- 




Figr. 105. 


Fig. 106. 


selves may be constructed with strong angle stiles, 
which are mitred together, and provided with either a 
hardwood tongue inserted in the joint, or with a 
wooden roll on the top to keep out the water. 

With skylights of this description, channels for con¬ 
densed water should always be provided. These are 
placed at the upper inner edge of the curb, the re¬ 
mainder of the inside face ot the curb being covered 
by either panelled framing or match boarding. 
























































PRACTICAL SOLUTIONS 


233 


Figs. 105 and 106 give details of a skylight having 
the form of a square pyramid. In this example the 
four triangular lights are mitred at the angles, and 
have wooden rolls over the joints. Figs, 107 and 108 
show elevation and part plan of a skylight with a 
curved roof surface. 

A Lantern Light differs from the skylights just de¬ 
scribed in having, in addition, vertical Sidelights. The 
sidelights consist of sashes, which, by being hinged or 



y u u u u 


Fig. 107. 



pivoted, are often available for ventilation. As they 
are in exposed positions, the greatest care is required 
in order to obtain watertight joints. When the side¬ 
lights are hinged on the bottom rail, they open in¬ 
wards ; when on the top rail they open outwards. When 
they are hung on pivots, the pivots are fixed slightly 
above the middle of the sash. Figs. 109 and 112 show 
details of a rectangular opening surmounted by a lan¬ 
tern light which is hipped at both ends, and has side¬ 
lights arranged to open inwards. 
































234 


MODERN CARPENTRY 


The construction of skylights and lantern lights af¬ 
fords good examples of the application of geometry 
to practical work as described in previous pages. When 



Fig. 109. 

the roof-lights are pyramidal as shown in Figs. 106 and 
110, and a separate frame is constructed as shown in 




o 

Plan 

Fig. 110. 


Fig. 110, the methods of obtaining the lengths and 
bevels of the hip rafters are similar to those described 
for getting hip rafters. When the roof-lights mitre 



































































































































PRACTICAL SOLUTIONS 


235 


against one another, the sizes of the lights and the 
bevels of the angle stiles which mitre together are ob¬ 
tained as shown at X in Fig. 112. With lights of curved 



Fig. 111. 

outline, the shapes of the hip rafters or angle-stiles, as 
well as the developed surfaces, are obtained as ex¬ 
plained before. 

Lay-Lights. At the ceiling level of roof-lights used 
for staircase wells, or in similar positions, it is often 



Fig. 112. 



Fig. 113. 



Fig. 114. 


considered advisable, for the sake of appearance, to 
have a horizontal second light called a lay-light. This 
consists of a sash—or if the space is large, a number of 


















































































































































236 


MODERN CARPENTRY 


sashes—fixed into frames in the ceiling. The chief 
feature of lay-liglits is in the attempt at decoration by 
arranging the bars in some ornamental design (Figs. 
113 and ]14). The lay-lights are often glazed with 
ornamental glass, which, although it improves the ap¬ 
pearance, diminishes the amount of light transmitted. 

Greenhouses and Conservatories. In this type of 
building which is largely constructed of wood and 
glass, the framework is usually of moulded and re¬ 
bated quartering, with side sashes fixed in the rebates. 
As in the case of skylights, the roof-lights, which in 
this case reach from the ridge to the eaves, have no 
crossbars, since these would impede the flow of water 
running down the slope of the roof. Care should be 
taken to have the bars strong enough to carry the glass 
without sagging; and it is well to remember that when 
a roof is of flat pitch a heavy snowstorm will throw a 
large additional weight upon it, while with a steep 
roof' the wind has much power. The distance apart of 
the bars which carry the glass ranges from 12 to 18 
inches, and the lengths of the sheets of glass should be 
as great as possible, so as to diminish the number of 
cross-joints, since these allow of accumulations of dirt 
which cannot be removed easily. These roof-lights are 
constructed in exactly the same manner as skylights; 
they are, however, often much larger, and require to be 
thicker, unless purlins are placed to support them. 
When, as is often the case, part of the roof-light is 
made to open, this part—often a narrow strip at the 
highest part of the roof (Fig. 115)—is made as a sep¬ 
arate light, which overlaps the upper edge of the fixed 
lower light. Additional ventilation is secured by ar¬ 
ranging the side sashes to open. 


PRACTICAL SOLUTIONS 


237 


The above description is intended merely to outline 
the broad principles of the construction of conserva¬ 
tories, but it should be remembered that the details, 
while conforming to casement and roof-light construc¬ 
tion generally, lend themselves to considerable varia¬ 
tion in design and arrangement. 



Fig. 115. 


A Bay Window with solid frame and casement lights 
is shown in Pig. 116 to 118. Two methods of fitting 
such a window, with folding shutters, are given in 
this plan. In the half plan at C, the shutters fold 
into a boxing projecting into the room, and at D they 
fold back upon the face of the wall, which is splayed 
to receive them. The sills of the frame are mitred at 








































































































238 


MODERN CARPENTRY 


the angles, the joint cross tongned and fixed with a 
handrail bolt, which should be painted with red lead 
before insertion. The joints in the head are halved 
together, the mullions stub tenoned and fixed with 



coach screws. The jambs are tenoned and wedged into 
the head and sill. The transom tenoned into the jambs 
and mullions, and secured with bolts. The mullions 
may be worked in one piece as shown at D, or built 
up as at C, and tongued and screwed together. 



































































































































































































































































































































































































































240 


MODERN CARPENTRY 



A Cased Frame Bay Window is shown in the half 
plan, half inside elevation, and central vertical section, 
Nos. 1, 2 and 3, Fig. 119. This window is composed of 

















































































































































































































































































































PRACTICAL SOLUTIONS 


241 


three ordinary sash frames, the sills connected at the 
angles either bv halving and screwing or by mitreing 
them and fastening the joint with a hand rail bolt. The 
heads are tied together with a short piece of 1 in. stuff 
screwed across the top of the joints, and the joints in 
the linings are covered by the mullions of the blind 
frame B. The latter, made 2 in. wide, forms an en¬ 
closure for Venetian blinds. Boxings are formed in 
the elbows between the sash frames and the interior 
face of the wall, the front of the opening being fin¬ 
ished off with a moulded ground and architrave. These 
form receptacles for the folding shutters, which are 
curved in plan, and when opened out convert the octag¬ 
onal bay into a segmental niche. The window back 
and the seat beneath are also curved to parallel sweeps. 
The window board also follows the sweep, and is re¬ 
bated to receive the shutters, a shaped bead being 
fixed on the soffit to form a stop at the top. Nos. 4, 5 
and 6 on the same plate illustrate another method of 
finishing a bay window. In this case the frame is solid, 
and is fitted with outward opening casement lights. A 
blind frame is provided, and the shutters fold on to the 
face of the jamb and wall, the outer edges passing be¬ 
hind the rebated edges of the architrave; the lat¬ 
ter is continued down to the floor, and elbow linings to 
correspond with the shutters are fitted beneath the win¬ 
dow board. These are fitted to the window back in 
the manner indicated by the dotted lines in the plan No. 
4. The section No. 6 shows the treatment of the roof 
of the bay, which is segmental in section and covered 
with shaped pan-tiles. The ribs, which are elliptic at 
the hips, are notched into a wall plate resting on the 


949 


MODERN CARPENTRY 





Fie:. 121 















































































































































































PRACTICAL SOLUTIONS 243 

stone cornice, and are nailed at tlie top into a shaped 
rib fixed on the face of the wall; the ribs are covered 
with weather boarding, which affords a good fixing 
for the tiles. The wall is carried by a breastsummer 
formed of two 12 in. by 6 in. balks bolted together, 
with spacing fillets between, and the soffit is carried by 
three brackets fixed to the breastsummer and the head 
of the window frame. 

A Pivoted Light in a solid frame is shown in eleva¬ 
tion in Pig. 122 and section Fig. 123. These are used 
chiefly in warehouses, lanterns, and other inaccessible 
positions, the lights being opened and closed either by 
cords and pulleys or by metal gearing. For small 
lights the frames are usually made out of 41/2 in. stuff 
by 2 or 3 in. wide. Small lights are pivoted horizontal¬ 
ly, large ones vertically. The pivot should be fixed to 
the frame, not the sash, and from % in. to 1*4 in. above 
the centre, according to the weight of the bottom rail. 
The lower part of the sash should exceed in weight the 
upper part, just sufficient to keep it closed; its action 
may be easity demonstrated by inserting two bradawls 
in the stiles, and balancing them on the fingers. The 
sash is inserted and removed from the frame either by 
means of plough grooves in the edges of the stiles, as 
shown by the dotted lines in Fig. 122, or by cutting a 
notch through the face of the stile for the passage of 
the pin, which is concealed when in use by the guard 
beads. This latter is the better method, as it does not 
reduce the strength of the sash, as does the former, by 
cutting away the wedging. The stop beads at the 
sides are cut in two, one part being fixed to the frame, 
the other to the sash. Their joints can be at any angle 






244 MODERN CARPENTRY 


























































PRACTICAL SOLUTIONS 


245 


greater than that made by a line tangent to the sweep 
at the point of intersection a Fig. 124, but for the pur¬ 
pose of using the Mitre Block, they are generally made 
at an angle of 45 deg. A curved joint has no advan¬ 
tage over a straight one, except in being more expen¬ 
sive. 

To Hang the Sash. Insert the pivots in the frame 
quite level, but do not screw them. Then with the try 
square resting on the top of the pins, square lines 
across the jambs. Then remove the pivots and insert 
the sash, which should be fitted rather tightly at first, 
and square the lines on to the sash. Return these on 
the edges, and keep the edge of the hole in the socket 
plate to the line, and the plate itself in the middle of 
the thickness. After the socket is sunk in, and the 
notches cut, test the sash and correct the joints, which 
should be a bare % in. clear all round. 

To Find the Position to Cut the Beads. After fit¬ 
ting them round, remove them and open the sash to the 
desired angle, which should be less than a right angle, 
so that the water may be thrown off. Lay the beads 
upon the sash upside down for convenience of marking, 
and draw a line along their edge upon the jambs at the 
point where the line meets the faces of the frame; 
square over lines as at a a Fig. 123; the position is 
shown in Fig. 122, the outer dotted lines indicating 
the beads. Next replace the beads and transfer the 
marks to them, cutting them off the mitre block (re¬ 
member that the mark is the longest point of the mitre). 
The upper portions outside and the lower inside are 
fixed to the frame, and these are shaded in the draw¬ 
ings. The remainder of the beads are fixed to the sash. 


246 


MODERN CARPENTRY 


The above describes the method when the sash is 
grooved. Where the beads are slotted, a variation 
must be made with reference to the top cut (see Fig. 


Sbv ///////////& 

m 

f 

! 

ft 

i 

_/ 

l 

/ J 

\ 

/ 

a 

i 

• -^ - 


» 

X. 

X 

,iy 

1 







Fig - . 125. 


124). In this case the sash must be drawn out and 
rested upon the pin, then the bead laid on it and 

marked as before, the intersection a giving the mitre 
point. 
























































PRACTICAL SOLUTIONS 


247 


A Bull’s-Eye Frame with a pivoted sash is shown in 
Figs. 125 and 126 and enlarged detail of the joint. This 
frame is built up in two thicknesses, glued and screwed 
together, each ring being in three pieces breaking joint. 
The beads may be steamed and bent round, or worked 
on the edge of a board that has been cut to the sweep, 



and cut off in two lengths. The sash is made in three 
pieces, with butt radial joints bolted together. To en¬ 
able the sash to open, a plane surface must be provided 
at the centre, ecpial to the thickness of the sash and 
beads, as shown by the dotted lines. Having fitted the 
sash in, and the beads around each side, brad them tern- 







































248 


MODERN CARPENTRY 


porarily to the sash, lay a straight-edge across it paral¬ 
lel with the centre, and square up with the set square 
a line at each side equal in length to the thickness, then 
cut the pieces so marked off with a fine saw, both beads 
and sash, and glue them to the centre of the frame, and 

hx the pivots to these frames and proceed as in a square 
irame. 



Laying Out a Circular Louvre: Suppose the frame 
of the louvre to be formed of four pieces, as shown at 
Mg. 127. These sections may be formed of one thick 
piece of plank, or may be built up of several thick¬ 
nesses. If of one thickness, the joints may be held to¬ 
gether with handrail screws, or dowelled and keyed. If 
of several thicknesses then the joints can be broken or 
overlapped, and the pieces either screwed or nailed 
together. To set out the louvre boards, make a sketch 
of the whole thing, full size, as shown at Fig. 127, then 
set up the section or side as shown at Fig. 128,' then 
project from the quick of the bead, then draw ’a line 
CD, Fig. 129, parallel to the inclination of bevel of 

























































PRACTICAL SOLUTIONS 


249 


the louvre boards. This will give the major axis of an 
ellipse. Bisect this line and draw E, F, G, at right 
angles to C, D, making E, F, and F G each, equal to 
half the radius from the centre of the frame to the 
quick of the bead. The ellipse C, E, D, G, may be 
struck by any of the methods shown. As, of course, all 
the louvre boards are at one angle, each forms a por¬ 
tion of the same ellipse. From the centre of the front 
edge of each louvre board, project across to the major 
axis as shown. Now from any of the louvre boards set 
off, at right angles, lines from the upper and lower 
surfaces as shown at H, Fig. 128, then the distance K, 
is the amount of projection of the lower surface in 
front of the upper. Taking half the distance of K, 
measure it off on each side of the centres 1, 2, 3, 4, 5, 6. 
Fig. 129, then through the points last obtained draw 
lines parallel with the minor axis, the upper line rep¬ 
resenting the top front board or arris of the boards, and 
the dotted line the bottom arris. Now measure the dis¬ 
tances K, from C to L, and from D to M, and construct 
the ellipse for the underside of the boards. The fol¬ 
lowing will be found a simple method for making out 
each louvre board: Cut a thin mould equal to one- 
quarter of the ellipse, the edges of the louvre boards 
having been planed to proper bevel, as shown at N, Fig. 
128, square the centre line across the piece that has 
to form the louvre board and lay it on the setting out at 
Fig. 129, the centre line of course corresponding with 
C D. The quarter mould which was prepared can now 
be laid on the louvre board, with its curve standing 
directly over the development, Fig. 129, the face side 
of the material being of course the side marked out. 


250 


MODERN CARPENTRY 


The curves for the underside may be marked off on the 
arrises direct from the development, and the mould 
then applied to the other side, taking care to adjust it 
in the right position and to the marks made. The ends 
should be sawed and turned to the lines. The next 
proceeding will be to set out the frame for the grooves; 
these are represented in the conventional sketches, 4, 5 
and 6. It will be noticed from Fig. 132 that the bottom 
louvre board is not grooved in all round; this is a bet¬ 



ter method than bringing it out to the front and thus 
destroying a part of the margin bead of the frame. The 
cutting of the grooves and the fitting in of the louvre 
boards requires careful working in order to get good 
joints. It must be clearly understood that Fig. 129 is 
not a full development of all the boards edge to edge, 
as that could not be represented in this space, but 
enough is shown to give a clear idea of how the lines 
are obtained. The full breadth of each is represented 
by the dimension lines O, P, R, S, T, U, V, and from this 
all the other is easy. 



















PRACTICAL SOLUTIONS 


251 


The Construction of Doors. Doors are named in ac¬ 
cordance with their modes of construction, position, 
style or the general arrangement of their parts, and 
also the method in which they are liung, as Battened, 



B 


Battened Framed and Braced, Panelled or Framed, 
Entrance, Vestibule, Screen, Sash, Diminished Stile, 
Double Margin, Gothic, Dwarf, Folding, Swing, Jib, 
Warehouse Hung, &c. The essentials in the design and 
construction of doors are, for the first, that they shall 
have a due proportion to the building or place they 




















































































252 


MODERN CARPENTRY 


have to occupy and be suitably ornamented; in the sec¬ 
ond, that their surfaces shall remain true and their 
parts be so arranged and connected that their shape 
will be unalterable by the strains of usage and the ef¬ 
fects of weather. The various example illustrated will 
indicate the points to be considered in designing doors 
foi sundry situations, and the methods of construction 
herein described will supply the necessary informa¬ 
tion to meet the constructive requirements. 

Battened and Battened Framed and Braced Doors 
are shown in Figs. 133 and 134. These doors are suit¬ 
able for positions where one or both sides are exposed 
to the weather. Little or no attempt is made to orna¬ 
ment them—economy of cost, strength and utility be¬ 
ing the chief requirements of this class of door, which 
ai e fitted to coach-houses, "W . C. ’s and outhouses gen¬ 


erally, 


The plain Battened Door (Fig. 133) is composed of 
battens A, from to 1 ] 4 in. thick, ploughed and 
tongued in the joints with straight tongues which 
should be painted before insertion, nailed to three 
ledges, B from 1 in. to 1% in. thick, usually with 
wrought nails long enough to come through and be 
clinched on the back side. The ends of the ledges are 
better fixed with screws, and their top edges as well as 
those of the braces C should be bevelled to throw off 
the water, as shown in the detail, Fig. 135. The lower 
edges may be throated or bevelled under, as shown. 
The braces should be placed so that their lower ends 
are at the hanging side, for if in the opposite direc¬ 
tion, they will be useless to prevent the door racking. 

1 heir ends should be notched into the ledges about 1 in. 



PRACTICAL SOLUTIONS 


253 


deep and 1% in- from the ends, with the abutment 
square to the pitch of the brace. Narrow doors are 
sometimes made without braces, but they seldom keep 
“square.” These doors are hung with wrought-iron 
strap hinges called cross garnets, which should be fixed 
on or opposite the battens, whether placed on the face 
or back of the door. 



Fig. 136. 


Fig. 137. 


The Framed Battened and Braced Door (Figs. 136 
to 137) differs from the former in that the battens and 
ledges are enclosed on three sides by a frame of a 
thickness equal to the combined thickness of the bat¬ 
tens and ledges, so that it is flush on each side with 



































254 MODERN CARPENTRY 

them. The boards are tongued into the frame at the 
top an <l sides, and the ledges are framed into the stiles 
with barefaced tenons. The braces should not be taken 
into the angle formed by the stile and rail, but be kept 
back from the shoulder about 1 in., as shown. If the 
brace is placed in the corner, the strain thrown on it has 
a tendency to force off the shoulder, unless the door is 
very narrow, when the brace will be nearly upright. 
These doors, as in fact all framed work exposed to 
damp, should be put together with a quick drying paint 
instead of glue in the joints, because ordinary glue has 
such an affinity for water that it will soften in damp 
situations releasing its hold, and also be the means of 
setting up dry-rot in the timber. The battens in these 
doors should be made 1/16 in. slack for each foot of 
width to allow for subsequent expansion, or otherwise 
the shoulder will be forced off. The framework of 
these doors is first made and wedges up, then the bat¬ 
tens folded in and driven up into the top rail and 
nailed to the ledges, after which the braces are cut 
tightly in and nailed to the battens in turn, and the 
whole cleaned off together. In large gates of this de¬ 
scription it is usual to stub tenon the braces into the 
rails, in which case they must be inserted first and 
'wedges up with the framing. 

Framed or Panelled Doors are of several kinds, dis¬ 
tinguished by the number or treatment of their panels, 
or by the arrangement of the mouldings, as follows: 

Two to Twelve Panel Doors. 

Square and Sunk. When a thin panel is used with¬ 
out mouldings, as shown at A in the elevation dia¬ 
gram of a Framed or four-panel door (Fig. 138). 


PRACTICAL SOLUTIONS 


255 



Fig. 138. 



Fig. 139. 


Moulded and Square. When one side of the panel is 
moulded and the other plain, as at B. 

Bead Flush. When one side of the panel is flush, or 
nearly, so, with the frame, and with a bead worked 
round the edges to break the joints, as at C. 










































































































256 


MODERN CARPENTRY 


Bead Butt. A\ hen the bead is worked only on the 
two sides of the panel, as at D. 

Raised Panel. When the centre part, of the panel is 
thicker than the margin. There are four varieties of 
raised panels: 



Fig. 140. 


1. The Chamfered. In this the panel is chamfered 

down equally all round, from the centre to the edge 

when square, or from a central ridge if rectangular as 
shown. 


k 



Plan and Sections of a Chamfered 
Raised Panel. 


Fig:. 142. Pig. 141 . 


2. Raised and Flat or Raised and Fielded. When a 
chamfer is worked all round the edge, leaving a flat 
in the centre, as at A, Fig. 142. 

d. Raised, Sunk and Fielded (as at B, Fig 142) 

When the chamfer starts from a marginal sinking be¬ 
low the face. 























PRACTICAL SOLUTIONS 


257 


4. Raised, Sunk and Moulded (as at C, Fig. 143). 
When the edge of the sinking is moulded. 

Stop Chamfered. When the edges of the framing are 
chamfered and stopped near the shoulders. 

Bolection Moulded. When the panel moulding stands 
above, and is rebated over the edges of the framing, as 
shown at Fig. 142. 




Double Bolection Moulded. When the moulding on 
each side of the door is made in one solid piece, grooved 
to receive the panel, and is itself grooved and tongued 
into the framing. This variety is shown at Fig. 142. 

Constructive Memoranda. The outside vertical mem¬ 
bers of doors (in common with all framed work) are 


























































258 


MODERN CARPENTRY 


called stiles. The one the hinges are fixed to is called 
the hanging stile, the one containing the lock the strik¬ 
ing stile. In a pair of doors the two coining .together 
are called the meeting stiles. The inside vertical mem¬ 
bers are the mountings, or more commonly muntings. 
The horizontal members are rails, respectively, top, 
frieze, middle, or lock, and bottom. The panels are 






Fig-. 145. 


named similarly. When the grain of these run hori¬ 
zontally, they are said to be "laying’’ panels; when 
vertically, "upright.” Doors are called Solid-Moulded 
when the moulding is wrought or "stuck” in the sub¬ 
stance of the framing itself, as is shown at Fig. 145* 
and Planted when the moulding is worked sep¬ 
arately and bradded around the frame, as shown at 

a, Fig. 143, and D, F ig. 143 1 / 4. These are also called 
sunk mouldings. 














































































PRACTICAL SOLUTIONS 


259 


Bead Flush Panels are commonly made as shown at 
a, Fig. 146, but such panels will, unless made of 
thoroughly seasoned stuff, inevitably split when dry¬ 
ing. The correct way to obtain the effect of bead flush 
panelling is to work the beads upon the edges of the 
framing, as shown in Figs. 147 and 148. 



Bead Butt Panels are better kept about 1/32 in. be¬ 
low the framing, as a truly flush surface is difficult to 
prepare through the yielding of the panel, and when 
produced, will seldom last, as the shrinkage of the 
panel and frame is unequal. Planted-in “sunk’ 7 mould- 

























































































260 


MODERN CARPENTRY 


mgs should be fixed to the framing, not to the panels, 
as shown in Fig. 143, D; for if fixed to the panels, when 
the latter shrinks the moulding will be drawn away 
from the frame, leaving an unsightly gap. The back 
edge of the moulding should be bevelled under as 
shown, so that when bradded in, the front edge will 
keep close down to the panel. As it is not permissible 



to brad polished mouldings, except in the case of in¬ 
ferior work, these are usually glued to the frame, and 
their back edges should of course be square. The panel 
should be polished before the moulding is planted in, 
so that in case of shrinkage a white margin will not 
be shown. When, however, the moulding is wide and 
thin, it is unavoidable that it be fixed to the panel to 
keep its front edge down, and to overcome the diffi- 
























































































PRACTICAL SOLUTIONS 


261 


culty of shrinkage. The stiles in vertical and the rails 
in laying panels are prepared Avith a second shallow 
grooA’e as shoAvn, and the moulding, made with an extra 
wide quirk, is cut in and pressed back into the groove. 
It is glued to the panel, and shrinks with it without 
showing a gap. The cross pieces across the ends of the 
panel have their quirks shot off to the proper width, 
and are sprung in after the side pieces are in place. 
These are dowelled but are not glued, or when there is 
a moulding at the back, the face side is slot screwed. 

Bolection Mouldings are intended to be fixed to the 
panels, which is rendered necessary by their great 
width and thickness, and they are rebated over the 
edge of the framing to prevent the interstices produced 
by the shrinkage showing. From % to 3/16 in. is suffi¬ 
cient for this purpose, and more should not be given or 
the edge will be liable to curl off. To prevent the panel 
being split by the fastenings when it shrinks, the 
moulding is fixed by slot screAvs, as shown in Fig. 143, 
the slots being cut across the grain. See the back 
vieAV, Fig. 144. The moulding should also be screwed 
together at the mitres, and the latter may be grooved 
and tongued with advantage, and dropped into the 
panel as a complete frame, where it is fixed as de¬ 
scribed, the heads of the screAvs being covered by the 
interior moulding, AA T hich, if a sunk one, is glued to the 
frame, and if a bolection, dowelled to the panel, or, as is 
sometimes done in inferior work, bradded, and the 
holes filled up with colored stopping. It is not good 
construction to glue the moulding to the panels in any 
case, as the alteration of size in the latter, due to the 
state of the atmosphere, is very liable to cause them to 


262 


MODERN CARPENTRY 


split, if fixed immovably, or when swelling, to disar¬ 
range the mitres. The bradding in of mouldings is less 
likely to do this if the brads are not placed too thickly, 
as they yield slightly to the pressure. Framing is 
usually grooved p 2 in. deep for the panels, and the lat¬ 
ter given y s in. play sideways, but fitting close length- 
vays of the grain. This is sufficient for panels up to 
2 ft. wide. Over that width the grooves should be % 
m* de <Tb and the panel enter % in. at each side. Ordi¬ 
nary dry stuff will eventually shrink about % in. to 
the foot, and will swell equally if exposed to damp. 
When wide panels are used they will warp less if glued 
up in several pieces, as the pull of the fibres is lessened 
by the cutting, and the effect of the warping is dimin¬ 
ished in the same ratio as their width. Much can be 
done to ensure the permanent flatness of panels by pay¬ 
ing attention to the way the boards have been cut from 
the tree. The direction of the annual rings on the end 
will indicate this, and the various pieces should have 
their similar sides placed together. What is meant by 
this will be rendered plain by an examination of Fig. 
149. When a panel is glued up with the hollow or heart 
sides of the rings all on one face as at A, and the board 
warps; it will case in one continuous curve, as shown in 
the unshaded diagram, whilst if glued up with the heart 
sides reversed alternately, as shown at B, it will as¬ 
sume the serpentine shape shown in the unshaded dia¬ 
gram. Boards cut radically or with the annual rings 
perpendicular to the surfaces, as at C, will swell less 
than the others, and will not warp perceptiblv 
Proportions. The size of doors depends so much upon 
the scale and design of the buildings they occupy, that 


PRACTICAL SOLUTIONS 


263 


no definite data can be given, within reasonable lim¬ 
its, for important doors ; but it may be pointed out that 
very large doors not only tend to dwarf a building or 
a room, but they also take up a great deal of space in 
opening, and the difficulty of preserving their accurate 
fitting increases in direct ratio with the size. The 
following may be taken as an indication of the more 
rflsual dimensions given to ordinary good class dwelling 
house doors: Entrance Doors, from 7 ft. to 8 ft. 6 in. 



illustrating the Effect 

. V 

of Position on the Parts 
of a Panel. 

Fig-. 149. 

high by from 3 ft. to 4 ft. 6 in. wide, 2 to 2% in. thick. 
Reception Rooms, 7 ft. by 3 ft. 3 in. by 2 in. Bed- 
Rooms, 6 ft. 8 in. by 2 ft. 8 in. by T>/ 2 in. Details of in¬ 
terior doors; stiles and top rails, in common work, out 
of 4Vi in., muntings and frieze rails 4 in., middle and 
bottom rails 9 in. Superior doors vary much, but gen- 





















































264 


MODERN CARPENTRY 


erally stiles and rails are somewhat wider than the 
above, muntings and frieze rails narrower. Height of 
lock rail usually 2 ft. 8 in. to its centre. This is a con¬ 
venient height for the handle, which is generally placed 
in the middle of depth of rail. When an entrance door 
is approached from a step the middle rail is kept about 
6 in. lower, to bring the height of the handle con¬ 
venient. 

Common Doors, both internal and external, are made 
of “yellow pine” or Georgia pine throughout. A better 
class of interior doors have yellow pine frames and 
white pine panels. The latter wood should not be 
used for external work, as it is far too soft and will 
not stand wet. Superior Internal Doors are made 
throughout of Honduras mahogany, black walnut and 
oak; also of pine and baywood, veneered with Spanish 
mahogany. External Doors of oak, teak, walnut and 
pitch pine. 

In constructing doors of any of the above mentioned 
figure woods, great attention must be paid to the ar¬ 
rangement of the members, so as to balance the figure, 
and this may also well be studied in the conversion of 
the plank. For instance, two stiles, each having pro¬ 
nounced figure at one end, and the other end plain, 
should have the figured ends placed at the bottom. This 
gives the effect of solidity, whilst the reverse would 
make the door look top heavy. Similarly the upper 
rails should be plain, the lower figured. It must be 
understood the above only applies when the wood is a 
mixed lot. When the wood is handsomely figured 
throughout, the point of most importance is the effect 
of its position upon the figure, and this is so great that 


PRACTICAL SOLUTIONS 


265 


in some of the light-colored woods, wainscot and bay- 
wood for instance, a piece that in one position will ap¬ 
pear richly figured will in another show quite plain 
and dull. The best way to judge the effect is to prop 
the pieces up in the approximate positions they will oc¬ 
cupy when finished, facing a top light. Then when 
standing a few feet off, the play of light on the fibres 
will be observed. Deep-colored woods, such as teak 
and Spanish mahogany, may have their figure brought 
out by slightly oiling them, which will facilitate their 
arrangement. Panels also require balancing, the more 



heavily marked ones being placed below plainer ones, 
and symmetrically arranged, either in pairs, or figured 
in the centre and plain outside. In all cases where the 
figure is coarse, taking a truncated elliptic shape, the 
base or wider part should be kept downwards. The 
panel at B is upside down from an artistic point of 
view. This arrangement is known in the workshop as 
“placing the butts down,” although as a matter of 
fact the width of the figure is not due to its being 
towards the butt end of the tree, but merely to the ac¬ 
cidental position the surface of the board occupies with 
relation to the annual rings, which are more or less 




























266 


MODERN CARPENTRY 



QODniMMLM 





*. 

7 


7 



\_:_ 

/ 









L 


N 




\ 







\ 


7 



\ 


/ 






i 



1 


N 



l 


V 





7 - 

7 



\ 


/ 









/ 


\ 



/ ■ 

\ 





4 



3 


Fig. 151. 


Fig. 152, 
















































































































































PRACTICAL SOLUTIONS 


267 


waved in length, due to crooked growth, and the board 
passing through them in a plane, their edges crop out 
on the surface in irregular elliptic shape. 

Double Margined Doors are wide single doors framed 
to appear as pairs of doors. They are used in openings 
too wide proportionate^ for a single door, but where 
half the opening would be rather small for convenient 
passage. Figs. 150 to 154 show elevation and sections 
of a Double Margined Entrance Door, typical of the 
style in vogue in the latter part of the eighteenth and 
early part of the nineteenth centuries. 

These doors are made in two ways. In the earlier 
method the central imitation stile, which in this case 
is really a munting, is made in one piece and forked 
over the top and bottom rails, which are continuous. 
The intermediate rails are stubtenoned to the central, 
and through tenoned and wedged to the outside stiles, 
but unless the stub tenons are fox-wedged, the shoul¬ 
ders are very liable to start, for which reason the 
method of construction now to be described is generally 
preferred. The door is composed of two separate 
pieces of framing, each complete with two stiles and a 
set of rails that are tenoned through and wedged up. 
The two portions are then united by a ploughed and 
tongued and glued joint, which is hidden by a sunk 
bead in the centre, as shown in the diagram, Fig. 155, 
and the parts keyed together with three pairs of hard¬ 
wood folding wedges. The door is sometimes further 
strengthened by having flat iron bars sunk and screwed 
into the top and bottom edges. The actual process of 
putting the door together is as follows: After the vari¬ 
ous rails and panels have been duly fitted and marked. 


268 


MODERN CARPENTRY 


each leaf is taken separately and the stiles knocked on. 
The one intended for the meeting stile haying been 



Figr. 154. Fig:. 155. Fig. 156. 

glued, is cramped up and wedged. Then the meeting 
stiles are made to a. width, grooved, jointed, and rebated 



































































PRACTICAL SOLUTIONS 


269 


for the beads, as shown in the detail, Fig. 150. The 
ends of the tenons and wedges should be cut back y 8 in. 
to prevent them breaking the joint when the stile 
shrinks. The mortises for the keys will have been 
made when the mortises for the rails were done, and 

> S 

cross-tongues are next glued in, the joint rubbed, the 
two stiles pinched together with handscrews, and the 
oak keys, well glued, driven in. At this stage the frames 
are stood aside to dry, after which the projecting ends 
of the keys are cut off, the panels inserted, and the two 
outside stiles glued and wedged in the usual manner. 
After the door is cleaned off, the grooves to receive the 
beads are brought to their exact size with side rabbit 
and router planes. Should iron bars be used, they are 
inserted in grooves made after the door has been shot 
to size. The bars should be about % in. shorter than 
the width of the door, so that the ends are not visible. 

Diminished stiles are sometimes cut out in pairs from 
a board or plank. When this is done, the back or out¬ 
side edge is shot straight and the setting out made 
thereon, the two portions being gauged to width also 
from the back, but the method more suitable for ma¬ 
chine working. Here the stile may be cut parallel to 
the full width, the face edge shot in the usual way and 
the setting out made upon that, the diminish being 
gauged from inside. In this method the mortises are 
made before the diminished part is cut out, to render 
that operation easier for the machinist. He should not, 
however, mortise right up to the sight lines on the di¬ 
minished part, because if the chisel is at all out of up¬ 
right when the waste is cut away, the mortise will be 
found beyond the sight line, which will be a serious 
defect. 


270 


MODERN CARPENTRY 


To Set Out the bevelled shoulders of the stile and 
rail. Taking the stile first, having as described in the 
first method gauged and faced up the inside edges, and 
set out the width of the rails and mortises on the back 
edge, square over on each side the sight lines of the 
middle rail as at a a, Fig. 156. Then draw a. second 
line representing the depth of sticking of the rail on 
the face side, and the rebate on the back side, as at b. 




Stile and Rail of a Diminished 
Stile Door. 

Method of Obtaining Shoulder Lines. 

Fig. 15C V 2 . 

Next run gauge lines down on each side of the dimin¬ 
ished part as working lines of rebate and sticking, and 
from the points of intersections of the stickings and 
rebates respectively, draw in the shoulder lines to the 
sight line of the lower edge of the rail at c. The onty 
difference to be made when the stile is prepared by 
the second method is that the sticking and rebate 

























PRACTICAL SOLUTIONS 


271 


gauges would be run from the original face edge in¬ 
stead of the actual diminished edge, and as before 
stated, the sight lines would be marked on the face edge 
instead of the back, and squared down to the intersec¬ 
tions. 

To Set Out the Rail. Mark on the bottom edge the 
“width” or sight lines of the widest part of the stiles. 
Square this point on to the upper edge as shown by 
dotted line in the sketch. Fig. 156, and set off there¬ 
from the amount of the diminish on the stile, as shown 
by the dotted line on the stile in the example: this is 2 
in. This line, knifed in on the edge, is the “sight line” 
of the upper part of the stile. Again set off beyond this 
line the amount of the sticking and rebate shown in 
the sketch by the lines e and f. Next run the sticking 
and rebate gauges en front and back sides, as shown at 
g, and square down the lines e and f to meet them. 
Then draw the shoulder lines from the intersections to 
the point d on each side. Having thus found the shoul¬ 
der line upon one rail, bevels may be set to them and 
used to mark any number. A contrivance sometimes 
used when a large number of similar shoulders have 
to be set out is shown in Fig. 157. This is known as a 
Shoulder Square. It consists of an ordinary set square 
provided with a movable fence or bar B, which is slot¬ 
ted to pass on both sides of the square, and is pivoted 
near the right angle. A set screw near the outer end 
of the bar passes through a concentric slot, and fixes 
the fence in any desired position. The pivot works 
tightly in a small slot to allow the lower edge of the bar 
to enter the right angle, and the outer edge of the 
square is also made a concentric curve to permit the 


MODERN CARPENTRY 



A Pair of Vestibule Doors. {Eighteenth Century Style.) 

Fig. 157. 







































































































































































PRACTICAL SOLUTIONS 


273 


easy passage of the end of the bar. The theory upon 
which the action of the tool is based is that the angle 
between the parallel bar B and either of the edges of 
the square is the complement of the remaining angle, 
the two combined forming a right angle which is the 
desired angle between the edges of a rail and stile. Its 
application is shown in the upper part of the figure, a 



An Adjustable Square 
for Bevel Shoulders. 

Fig. 157ha. 


being the rail, b the stile. Either a rail or stile is first 
set out as described above, the edge of the square set 
to the shoulder line, and the bar brought up to the 
face of the work and fixed with the set screw when it 

































































274 


MODERN CARPENTRY 


is ready for application to the other piece. The mould¬ 
ing upon a diminished stile should not be mitred but 
continued on to the shoulder, and the rail scribed over 
it, which will prevent an open joint occurring should 
the rail shrink. When doors, after knocking together, 
are stored for a second season, a slight difference will « 
have to be made in the setting out of the shoulders of 
the middle rail. The wider part of the stile will shrink 
more than the narrow part, and consequently if the 

shoulders are set out accuratelv at first as described 

«/ 

above, when they are refitted the shoulders will be 
found short at the lower ends. To prevent this, allow 
about 1/32 in. extra on the lower part of the shoulder 
at each end of the rail. 

A Gothic Door of the Tudor period is shown in the 
elevation in Fig. 158 and section in Fig. 159. The head 
is four-centred. The upper panels are pierced tracery, 
and the lower ones carved drapery. The mouldings 
in this type of door are invariably stuck solid, and 
those on the stiles stopped at the sight lines of the 
rails. The mouldings on the latter are also frequently 
stopped at the muntings as shown, especially in the 
earlier work. Many of the doors, however, of the Tu¬ 
dor period have the upper ends of the muntings mitred. 

In modern work of this style, w r hen the mouldings are 
not stopped, it is usual to scribe them at the intersec¬ 
tion. Chamfers, however, are always stopped to ob¬ 
tain a square-built shoulder for the munting, as a 
shoulder scribed over a chamfer soon gets faulty 
through the shrinkage. 

Mediaeval doors were always constructed of oak, 
but pitch, or Georgia pine is now much used in this 


PRACTICAL SOLUTIONS 275 

C 



D 

Fig. 158. Fig. 159. 


style of work. The older examples are mortised, ten¬ 
oned, and pinned together, wedging and glueing be¬ 
ing’ a modern invention. The joints at the head are 




























































































































































































































































































































276 MODERN CARPENTRY 

usually slip tenons, pinned, or with dovetail keys in¬ 
serted. These joints in a modern door would be se¬ 
cured either with a hammer-headed key or handrail 
bolts. 



Fig. 160. 


Fig. 160 shows the elevation of a superior five-pan¬ 
elled interior door with its finishings. Fig. 161 shows 
a vertical section through the opening, and Fig. 162 is 





































































































































































































PRACTICAL SOLUTIONS 


277 





a plan showing in outline the framed soffit lining. Fig. 
163 is an enlarged section of one stile and part of 
panel, mouldings, etc. 


Fig. 163. Fig. 162. 








































































































































































































278 


MODERN CARPENTRY 


Revolving Doors. —An arrangement of vestibule 
doors, suitable for banks, hotels, etc., is shown in plan 
in Fig. 164. The doors are arranged at right angles to 
each other, and revolve around a vertical axis like a 
turn-stile. Curved side frames, each a little wider than 
a quarter of a circle, are fixed on each side of the door¬ 
way. A suitable width for the doors is 3' 6". The ad- 



Plans of Revolving Vestibule Doors. 


Fig. 164. 


Fig. 


165. 


vantages of such an arrangement is that it is noise¬ 
less and draughtproof, the latter feature being obtained 
by having an india-rubber tongue fixed in the outer 
edge of each door. The doors are so hung that alter¬ 
nate doors can be folded back against the adjacent 
ones (Fig. 165), and thus give an uninterrupted pas¬ 
sage when required. 

Other Panelled Framing.—Framework filled in en¬ 
tirely with wooden panels, or with wooden panels in 
the lower part and glass in the upper part, is also re¬ 
quired in the fittings for offices, for school partitions, 















PRACTICAL SOLUTIONS 


279 


and for screens in churches, business premises, etc. The 
arrangement of the framing is similar to that of doors, 
and the same terms are used to describe the various 
parts, the only difference being the proportions of 
height and width; these are, of course, .governed by 
special requirements. 

Superior Doors.— In superior work, where the doors 
and surrounding framework are made of ornamental 
hardwood, it is often necessary to construct a door 
which shall be of one kind of wood on one side of the 
door and an entirely different kind on the other side. 
This would be necessary, for example, with a door 
opening from an entrance hall fitted entirely with oak 
into a room, the fittings of which must all be of walnut 
or mahogany. Such a door may be constructed in two 
thicknesses, each of the respective kind of wood, and 
each of a thickness equal to one-half of that of the 
finished door. The two parts are then secured together 
by tapering dovetailed keys, and the edges of the door 
are afterwards veneered to match the side of the door 
to which they correspond. Figs, 166 and 168 give de¬ 
tails of this kind of door. 

Grounds.— The architraves surrounding an opening 
are nailed to the lining, or where possible to the frame. 
In the best class of work, however, it is usual not to 
fix the door frames until the plastering is finished. 
Rough wooden battens or Grounds, of thickness equal 
to that of the plaster, are fixed to the walls around all 
door and window openings. These serve as a guide 
to the plasterer, and the door frames and the sur¬ 
rounding architraves are secured to them. When it 
is not desirable to have any nail holes visible in the 


280 


MODERN CARPENTRY 



Elevation 

Fig. 166. 


-£ Vertical Sectioni 

Fig. 167. 












Fig. 168. 







































































































































































































































PRACTICAL SOLUTIONS 


281 


finished surfaces, the door frames and architraves are 
fixed by screws. 

The fixing of the architraves around such a door¬ 
way affords a good example of Fixing by Secret Screw¬ 
ing. The mitres of the architraves are first glued and 
secured with dovetail keys or slip feathers. Stout 
screws are turned into the grounds about 12" apart, 
being left so that the head of the screw projects about 
haif-an-ineh in front of the surface. On the back side 




Fig. 169. 


of the architrave, exactly opposite the screw heads, 
small holes—equal to the size of the shanks of the 
screws—are bored; and about three-quarters of an 
inch below these, larger holes—of size equal to the 
heads of the screws—are bored. Each small hole is 
connected to the large one adjacent to it by a slot, the 
depth of which is slightly greater than the projec¬ 
tion of the screws. The architrave is fixed by placing 

















282 


MODERN CARPENTRY 


it against the wall with the larger holes fitting on the 
screws, and then carefully driving it down so that the 
heads of the screws hook into the fibres behind the 



Elevation 
Fig. 170. 



Vertical Section 
Fig. 171. 


slots. Bv placing the screws so that they are slightly 
inclined, the tendency is to draw the architraves closer 
to the wall. Fig. 169 shows the explanatory detail. 

The above remarks upon door frames, linings, etc., 
apply especially to the doors of dwelling-houses. Door 




























































































































































































































PRACTICAL SOLUTIONS 


283 


frames for warehouses, workshops, outbuildings, etc., 
do not as a rule require linings or architraves, a small 
fillet being nailed into the angle between the door 
frame and the wall instead. Vestibule doors are often 
hung to swing both ways, and the door frames have 
a hollow rebate or groove in the middle of the width 
of the frame, to receive the rounded edge of the door 
(Figs. 170, 171 and 172). Many of the heavier kinds 
of framed and ledged doors are not provided with 
wooden frames but are hung with bands and gudgeons, 
or arranged to run on pulleys as described elsewhere. 



Fig. 173 shows an ordinary sash door with three 
panels below. Fig. 174 shows a section of door and 
frame. Fig. 175 shows plan of door and part of cross 
section of frame. Fig. 176 shows the height and width 
rod of a door, which guides the workman in laying out 
his work. 

External Doors are invariably hung in Solid Frames. 
Internal doors, chiefly to build up casings or Linings, 
of comparatively thin substance. In certain positions, 
such as vestibules and shop-fronts, where there are no 
wall openings to line, solid frames are also used for 













284 


MODERN CARPENTRY 

























































































































































285 


PRACTICAL SOLUTIONS 




4 


Fig. 174. 


Height and Width Rod of a Door. 
Fig. 176. 










































































286 MODERN CARPENTRY 

interior doors. The members of solid frames are usu¬ 
ally made of square section or slightly thicker than 
wide; this arrangement may, however, be varied to 
meet the necessities of the design; the rebates, stops, 
mouldings, &c., are worked in the solid. The outer 
vertical members of these frames are called josts or 
jambs; interior ones, mullions. The horizontal mem¬ 
bers are sill, transom and head. The jambs are framed 
between the head and sill, chiefly that the ends or 
horns of the latter may run bevond the frame, and so 
provide fixings that can be built into the wall; and 



Fig. 177. 

also because the shoulders of the post form a better 
abutment for carrying any load that may be thrown 
on the head than the edge of a tenon would. Transoms 
are cut between the jambs and also between the mul¬ 
lions when these are used. 

A Segment-headed Frame is shown in Fig. 177, and 
enlarged details of the joints in Figs. 178, 179, 180. The 
heads of these frames are cut out of the solid when 
the rise will permit of their being cut from deals of 
ordinary width. When this cannot be done, they are 























PRACTICAL SOLUTIONS 287 

made in two lengths, jointed at the crown, and fas¬ 
tened with a handrail bolt. The horns are taken out 
level at the springing line, and the back is made 
roughly parallel to the shoulders for convenience in 
fixing. When the frame is 4 in. and upwards in thick¬ 
ness, double tenons should be used, as shown in Fig. 
179;. and if the position in which the frame is to be 
fixed does not admit of the horns being left on, the 
mortises should be haunched back, as shown in Fig. 



180, although the horns would not be cut off until the 
frame was fixed, as they would be required for the 
purpose of cramping up the frame. 

A Semi-headed Frame may have its head cut in two 
or three lengths and bolted together, but is frequently 
built up as shown in Figs. 181, 182, 183. The jambs 
and transom are worked solid, but the head is formed 
in two thicknesses glued and screwed together, one 





































288 


MODERN CARPENTRY 


layer being in two lengths, the other in three, so as 
to break joint. This is both a strong and economical 
way of forming a head, because the grain is less cut 
across than it would be in a head cut out of one thick¬ 
ness, and the labor of rebating is also dispensed with, 
the inner ring being kept back % in. to form a re¬ 
bate. The head is fastened to the jambs by hammer¬ 
head tenons and shoulder tongues, as shown in Fig. 
182, and double tenons are used for the transom to 




A Semi-headed Solid Frame. 



Fig. 181. 


Fig. 182. Fig. 183. 


avoid cutting the root of the head tenons. The tran¬ 
som is bettei kept about 3 in. below the springing as 
shown, to ensure a strong joint. But if the exigencies 
of the design necessitate its being placed at the spring¬ 
ing, then the jamb should be carried above the spring¬ 
ing, and a portion of the curve worked upon it, because 










































PRACTICAL SOLUTIONS 


289 


if the two joints are made together, the connection will 
be very weak. 

There are four varieties of door casings or Jamb Lin¬ 
ings as they are also termed, viz., Plain, Framed, Double 
Framed, and Skeleton Framed, these names defining 
the method of construction. Other sub-names are also 
used denoting the nature of the ornamentation, as in 
doors, with which they agree in the general arrange¬ 
ment of their parts. The term Plain is applied to any 
wall lining however it may be treated, if it is made 
of one fiat board or surface. 

A “Set” of Linings comprise a pair of jambs and a 
head or soffit lining. The fiat, against the edge of 
which the door rests when closed, is called the Stop, 
and in common work these are merely nailed upon the 
surface of the main lining, being kept back from the 
edge sufficiently far to form a rebate for the door. In 
better work the rebate is worked in the solid, the lin¬ 
ing in such case being thicker. Not less than l 1 /? in. 
stuff should be used for any lining to which a door 
• has to be hung, as the rebate takes y 2 in. out of the 
thickness, leaving only 1 in. for screw hold for the 
hinges. This, however, may be supplemented by hinge 
blocks glued to the back of the lining just behind 
where the hinges will be inserted, as shown on one side 
of the isometric sketch of a set of Plain Jamb Linings 
(Fig. 184). When the lining is rebated on both edges, 
it is said to be “double rebated.” Plain linings are 
not suitable for walls thicker than 14 in., in conse¬ 
quence of the amount of shrinkage which disarranges 
the finishings, and even in 14 in. work they are better 
framed. 


290 


MODERN CARPENTRY 



Rebated Set of Plain Linings with Double 
Set of Grounds. 

Fig. 184. 


A 































































PRACTICAL SOLUTIONS 


291 


Grounds. —Figs. 188 and 189 are light frames form¬ 
ing a boundary to all openings in plastered walls, their 





Head of a Framed Jamb Lining. 
Fig. 185. 


purpose being to act, as the name implies, as a ground¬ 
work for the linings, architraves, &c., and also as a 



Fig. 186. 

margin and gauge for the thickness of the plaster. 
They are either bevelled or grooved on the back edges 
to form a key for the plaster, and should be framed up 














































































292 


MODERN CARPENTRY 


perfectly square, and with faces and inside edges true, 
straight and square, to ensure good fittings in the fin¬ 
ishings. They are fixed by nailing to plugs or wood 



End of a Jamb Lining showing Tongues. 
Fig. 187. 


bricks in the walls, and should be secure, plumb and 
out of winding. They are usually prepared out of 1 in. 



Elevation and Edge View of a Set 
of Grounds. 

Fig. 188. Fig. 189. 

stuff, their width depending upon the width of the 
architraves, which should overhang them about % in. 
More cover than this is not advisable, or the fixing for 



































































PRACTICAL SOLUTIONS 


293 


the outside of the architrave will be lost, and for the 
same reason much bevel should not be given to the 
back edge. A % in. will provide quite sufficient key 
for the plaster, and also, as the grounds have to be 
bevelled before glueing up, a very thin edge is liable 
to be crushed in cramping them up. The sets of 
grounds on each side of an opening are connected 
across the jamb by 1 and 2 in. slips called Backing 
pieces, dovetailed, and nailed to the edges. Grounds 



Fig. 190. 


are, however, very often fixed flush with the brickwork, 
which does not permit of the use of backings; it is a 
very inferior method, the linings either touching the 
wails, or if made smaller, having to be fixed with fold¬ 
ing wedges and firring pieces, which do not afford so 
firm a fixing. When the architrave is of such width 
that the sides of the ground are required more than 
5 in. wide, it is not advisable to make them in one piece, 
because from their position they are very liable to cast, 
and so spoil the appearance of the finishings. In such 
cases, the grounds should be formed as skeleton frames 














































294 


MODERN CARPENTRY 


with rails and stiles, where required from 3 to 5 in. 
wide (see Fig. 190). 

Grounds are sometimes moulded or beaded on theii 
inside edges, as in Fig. 191. They are then called 
Architrave Grounds or Moulded Grounds, and are fixed 
to the edges of the linings, which in such cases must 


A Moulded Ground. 
Fig - . 191. 



Method of Arranging 
“ Grounds” for Glueing Up, 

Fig - . 192. 


be fixed first. The moulded ground is more frequently 
used in window and shutter openings than in door 
openings. All joinery finishings are fixed to grounds, 
but these are seldom framed, except in the cases of 
door, window and shutter openings, all other work 

being fixed to rough grounds, which will be illustrated 
in their appropriate place. 











































PRACTICAL SOLUTIONS 


295 


Framed Grounds are usually wedged up and sent 
out from the shop in pairs, as shown in Fig. 192, Fig. 
193 being an edge view to larger scale. Two sets are 
lightly nailed face to face, and with heads reversed, 
in which position they can be knocked apart and glued, 
then cramped, squared, and wedged up with facility. 

Architraves are the moulded borders or frames to 
window or door openings. When square at the back, 
as in Fig. 191, they are termed Single-faced; when 
moulded at the back, as in Fig. 185, Double-faced. No 



Framed Architrave 
and Grounds. 


Fig. 193. 


architrave should be made wider than 6 in. in one piece, 
as there is great danger of its splitting when shrink¬ 
ing, being fixed necessarily at both edges; but when 
this size is exceeded, the moulding should be made up 
of two or more members, grooved and tongued to¬ 
gether at some convenient point, as shown in Figs. 185 
and 193, the latter being an example of a very wide so- 
called Framed Architrave ; then, if fixed at both edges, 
it will be free to shrink in the middle. 




















296 


MODERN CARPENTRY 


Heavy Architraves are ploughed and cross tongued in 
the mitres, and in some cases framed or stub mortised 
and tenoned, as shown in Figs. 194 and 195. These 
are glued and fixed with screws turned in from the 
back. Small handrail bolts are also used to draw up 
the mitres in thick mouldings. Sets of architraves so 
treated are framed up on the bench, provided with 
stretchers at the bottom end, and braced to prevent 



An Architrave Tenoned and 
Mitred. 

Fig. 194. Fig. 195. 


racking during transit, and are fixed complete. Smaller 
ones are mitred and fixed piece by piece. In the best 
class of polished work, secret fixings are used for se¬ 
curing the architrave to the grounds. These are shown 
in Figs. 196, 197 and 198. A number of stout screws 
are turned into the grounds with their heads project¬ 
ing uniformly about *4 in. Corresponding holes and 
bevelled slots are made in the back of the moulding, 
which is dropped on the screws, and carefully driven 

































PRACTICAL SOLUTIONS 


297 


down in the direction of the arrow. The architrave 
should he driven on to the screws dry in the first place; 
two workmen being employed in the operation, one 
upon a scaffold carefully drawing down the architrave 



Elevation of Back of Architrave. —— Section of 
Architrave, and Elevation of Face of Grounds. 

Fig. 198. Fig. 197. Fig. 196. 


equally on each side; the other pressing the moulding 
tightly towards the grounds, and striking a piece of 
wood with a heavy hammer over each screw as his fel¬ 
low strikes the top. After being duly fitted the archi¬ 
trave may be knocked up again, and the front edges 
only lightly glued before replacing. 

Foot or Plinth Blocks are used at the bottom of all 
architraves in good work, and are secured in two ways 





















































298 


MODERN CARPENTRY 


(see Pigs. 199 and 200). The latter is the better way, 
as there is less difficulty in fitting the shoulder, and 
also less danger of splitting the plinth, when driving 
the dovetail tenon in; this is glued and secured with 



Method of Fixing Plinth to Architrave. 

Fig. 199. Fig. 2*00. 


a screw turned in the back. . A handscrew may be 
used to grip the sides of the block whilst driving the 
latter on to prevent its splitting. 

Door and Jambs Complete.—Figs. 201 and 202 show 
a door, with jambs and linings complete. The details 
for constructing and fastening in place are all shown. 
Each member of the door is named, also base and 
grounds. 

Figs. 202 to 220 show a number of typical doors, some 
of which will be found suitable for any particular pur- 
pose. 









































PRACTICAL SOLUTIONS 


299 



n_L£T 


MAUNCMINC 


Fig. 201 


Fig. 202 























































































































































































































300 


MODERN CARPENTRY 






n 

[ 







i 



J 


Fig. 203. 




Fig. 205. 



Fig. 206. 



Fig. 207. 

































































































































































































PRACTICAL SOLUTIONS 


301 




Fig. 210. 
























































































































































































































302 


MODERN CARPENTRY 








Fig. 218. 


Fig. 219. 


Fig. 220 













































































































































PRACTICAL SOLUTIONS 


303 



Fig. 221. 


Fig. 222. 



































































































PART III. 


MECHANICS OF CARPENTRY. 

In this part I intend to give the reader some prac¬ 
tical instructions in “The Mechanics of Carpentry/’ 
with the methods of calculating the strength of trusses, 
and timbers for structural purposes. All the rules 
shown have been taken from the best authorities, as 
Tredgold, Barlowe, Hurst, Riley, T’rautwine, Hatfield, 
Kidder, Nicholson and others. I have, in a few in¬ 
stances, corrected, and added to some of the propo¬ 
sitions, which I have found in practice, to be neces¬ 
sary. 


INTRODUCTION TO THE MECHANICS OF CAR¬ 
PENTRY. 

It is well known that some members of a framed 
structure must be made stronger than others. The 
reason is that the weights or other forces acting on a 
truss differ from each other in magnitude and direc¬ 
tion. It is obviously necessary, therefore, to be able 
to estimate the various forces acting, so that the mem¬ 
bers may be made of the required strength without 
undue waste of material. The general principles 
underlying the measurement of forces may with ad¬ 
vantage now be considered briefly. 

305 


306 


MODERN CARPENTRY 


The Nature of Force. Force may be defined as that 
which moves, or tends to move, a body at rest, or 
which changes, or tends to change, the direction or 
rate of motion of a body already moving. A familiar 
example of force is met with in gravitation, whereby 
an object has a tendency to fall to the ground. In 
order to support it, an upward force equal to the 
weight of the object must be exerted. The phrase 
“equal force” implies that forces can be measured. 
In this country they are usually measured in terms of 
weights in lbs., cwts., etc. Any one who lias seen a 
pulley, or lever, at work knows that the direction of 
application of a force can be changed. Evidently, 
then, forces can be represented graphically. Lines 
drawn to scale are employed and these can be ar¬ 
ranged to exhibit at the same time both the magnitude 
and the direction of the forces. Thus, a weight of 10 
lbs. acting vertically downwards can be represented 
by a vertical straight line 10 units in length. If the 
unit of length be Vs", the line will measure ten times 
; whereas, if the unit of force be represented 
by a length of 1 ", the graphic representation of the 
force will be a vertical straight line 10 " long. 

Resultant of two or more Forces. (1) When two or 
more forces together act at 9 point in the same direc¬ 
tion and in the same straight line, the resultant force 
is equal to the sum of the components. 

Example, (a) If two 10 lb. weights attached to a 
cord are hung upon the same nail, the resultant weight 
acting upon the nail is 10 + 10=20 lbs. 

(2) If two equal forces together act at the same 
point in opposite directions, but in the same straight 


MECHANICS OP CARPENTRY 


307 


line, they neutralize each other, and the forces are said 

to be in equilibrium. 

Example. A spring balance carries a weight of 6 
lbs. The index finger of the balance shows that the 
spring exerts an upward force equal to the downward 
force—the weight; and a state of equilibrium is ob¬ 
tained. 

If the two unequal forces together act at the same 
point in opposite directions, but in the same straight 
line, the resultant force is equal to the difference be¬ 
tween the forces, and is in the direction of the greater. 

It is evident, then, that the directions of the forces, 
and therefore the angles they make with one another, 
must be considered in determining the forces acting at 
any given point. 

If a flexible string be attached to a weight, and then 
passed .over a frictionless pulley, there will be the 
same tension in every part of the string, irrespective 
of any change of direction caused by using the pulley. 

To illustrate these facts clearly, suppose that two 7 
lb. weights, connected by a cord, hang over a smooth 
peg as shown in Fig. 1. The total weight on the peg, 
neglecting the weight of the cord (which may thus 
be any length), is 14 lbs., the sum of the two weights. 

Again, suppose three such pegs in a horizontal 
straight line, and the cord and weights to be passed 
over them as shown in Fig. 2. Evidently the weight 
on the central peg is nothing. Now, suppose the out¬ 
side pegs to be lowered slightly, as shown by dotted 
lines in the figure; the central peg will now carry a 
small proportion of the weight, and the more the out¬ 
side pegs are lowered, the more weight will be thrown 


308 


MODERN CARPENTRY 


on the central peg, until, as shown in Fig. 1, it carries 
all the weight, i. e., 14 lbs. Therefore the weight upon 
the central peg varies according to the direction of the 
forces acting on it—from nothing in Fig. 2 to 14 lbs. in 
Fig. 1. 

The magnitude and direction of the resultant force 
acting upon the central peg, and upon each of the out¬ 
side pegs, can be determined by the parallelogram of 
forces. 



at a point be represented in magnitude and direction 
by the adjacent sides of a parallelogram, the resultant 
of these two forces will be represented in magnitude 
and direction by that diagonal of the parallelogram 
which passes through the point. 





















MECHANICS OF CARPENTRY 


309 


Example 1. The angle at A, when the Cord passes 
over ihe pegs B 1? A,^, shown by the dotted lines in 
Fig. 2, is given. Determine by the parallelogram of 
forces the stress on the peg A, i. e., the single force 
acting through the point A, which shall be equal in 
effect to the forces AB X , AC X acting together. 

Produce AB 1 and AC X , and mark off on each line 7 
units, measuring from A. Then A1 and A2 represent 
in magnitude and direction the forces caused by the 
loads. Complete the parallelogram by drawing ID 
parallel to A2, and 2D parallel to Al. The length of 
the diagonal A D, measured in the same units as the 
lines Al and A2, represents the magnitude of the re- 
sultant force—i. e., the stress on the peg A. The di¬ 
rection of the force will obviously be downwards. A 
force represented in magnitude and direction by D A 
would evidently counterbalance the force A D, and 
would therefore counterbalance Al and A2 acting to¬ 
gether. Forces which balance each other are said to 
be in equilibruim. 

Example 2. Determine the magnitude and direction 
of the single force which will replace the two forces 
exerted by the cord and weight on the peg (Fig. 2). 

Draw ab 7 units long (Fig. 3) and parallel to the 
cord Al in Fig. 2. From b draw be also 7 units long 
and parallel to the cord below the peg B x . Complete 
the parallelogram by drawing dc and ad parallel to 
ab and be respectively. Then the diagonal bd gives 
the magnitude of the required force, the direction of 
which is from b to d. 


310 


MODERN CARPENTRY 


Example 3. Fig. 4 shows the application of the 
parallelogram of forces to determine the resultant 
force on the peg B. 

In the above examples no allowance has been made 
for the weight of the cord or for the friction on the 
pegs. It is assumed in each case that the forces are 
acting at the point of intersection of the straight lines 
produced. 




Example 4. Two forces of 10 and 6 lbs. respectively 
act from a point and in directions which are at right 
angles to each other. Determine the magnitude and 
direction of the single force which can replace the two 
forces. 

Let the line AB (Fig. 5) represent in magnitude a 
force of 10 lbs. acting at the point A in the direction 
indicated by the arrow, and A C a force of 6 lbs. acting 





MECHANICS OF CARPENTRY 


311 


at right angles to A B. Complete the parallelogram 
A C D B. Then the length of^tlie diagonal A D rep¬ 
resents the magnitude of the resultant force, and the 
direction in which it acts will be from the point A, as 
shown by the arrow. 

It must be understood clearly that a resultant is a 
force which can take the place of, and will produce 
the same effect as, two or more forces. To maintain 
equilibrium, the resultant force must be counterbal¬ 
anced by an equal force acting in the opposite direc¬ 
tion. The force so acting is called the equilibrant. 



Example 5. Figs. 6 and 7 show the magnitude and 
direction of the resultant force when forces of 9 and 6 
lbs. respectively act at angles of (a) 120°, (b) 45 . 

The simple apparatus shown in Fig. 8 clearly illus¬ 
trates the principle of the parallelogram of forces. On 
a vertical board are fixed two small pulleys by means 
of screws, so that they revolve with as little friction as 
possible. By making a three-way string, passing it 
over the two pulleys, and adding varying weights to 



312 


MODERN CARPENTRY 


each of the three ends of the string, it can be demon¬ 
strated clearly how the three forces act. In Fig. 8 the 
weights are respectively 5, 6, and 4 lbs. By drawing 
the parallelogram A B D C, such that A B equals 5 
units in length, and A C equals 4 units, the diagonal 
D A is found to measure 6 units, and to represent the 
magnitude of the middle weight. If other weights are 
attached to the ends of the strings, different results 
will, of course, be obtained. 



Triangle of Forces. The triangle of forces is used to 
determine the magnitude and direction of any three 
forces which balance each other. The rule may be 
stated as follows: If three forces acting at a point are 
in equilibrium they can be represented in magnitude 

and direction by the three sides of a triangle taken in 
order, 






MECHANICS OF CARPENTRY 


313 


Example 1. The forces acting upon A (Fig. 8) are 
in equilibrium. 

Since the length of the line A B=5 units, and the line 
B D is parallel and equal in length to A C=4 units, and 
the diagonal D A is in a line with the direction of the 
middle vertical weight and equal in length to 6 units; 
then the sides A B, B D, and D A of the triangle A B D 
represent both in magnitude and direction the forces 
acting at the point A. 



Fig. 8. 


To save confusion it is usual, however, to draw a sep¬ 
arate triangle to illustrate these forces. A somewhat 
different system of lettering also simplifies the consid¬ 
eration of the examples. This is known as Bow’s nota¬ 
tion. In it the two letters denoting a force are placed 
one on each side of the line representing the force, that 















314 


MODERN CARPENTRY 


is, in the spaces between such lines. Thus in Fig. 9 the 
three forces acting at the point o are referred to as 
A B, B C, C A respectively. 

Example 2. Given the magnitude (9 lbs.) and the 
direction (indicated by the arrow) of A B, and the 
angles which the directions of the three forces make 
with each other, it is required to find the magnitude 
and direction of B C and C A when the forces are in 
equilibrium. 



b 


Fig. 9. 


Fig. 10. 


Draw the line a b (Fig. 10) parallel to the direction 
of action of the force A B, 9 units long, and in the 
direction shown by the arrow. From b draw b c paral¬ 
lel to B C until it meets a c drawn parallel to C A. Then 
the triangle a b c is the triangle of forces, and the 
direction of the forces B C and C A can be found by 
taking the sides of the triangle in order, viz., a to b, 
b to c, c to a; and these directions give also the direc¬ 
tions of action of the forces represented by the lines 
parallel to a b, b c, and c a respectively. Thus A B acts 
from the joint o; BC acts towards o; and C A acts 
from o. 







MECHANICS OF CARPENTRY 


315 


The following examples show the application of these 
principles to simple practical questions. 

Example 3. A rope bears a tensile stress (pull) of 
30cwts. Find the magnitude of the stress in each of 
two other ropes which make an angle of 60° with each 
other, and together balance the stress in the first rope, 
supposing the second - and third ropes are equally 
stressed. 



Fig. n. 


Fig. 11 shows the application of the triangle of forces 
to the solution of this question, the answer giving the 
stress in each rope as 17.32 cwts. By going round the 
sides of the triangle in order, it will be seen that the 
force in each of the three ropes acts from the joint. 

Example 4. A buckling-chain is used to raise heavy 
blocks of stone. What is the amount of stress in the 




316 


MODERN CARPENTRY 


links of the chain when raising a weight of one ton, if 
the buckling-chain is: 

(a) pulled tightly as in Fig. 12; 

(b) placed loosely round the stone as in Fig. 13. 



The correct solution of this question depends on (1) 
the weight of the stone; (2) the angle between the 
forces A C and B C. 



The application of the triangle of forces in each case 
(Figs. 12 and 13) shows that the stresses AC and BC 
are more than twice as great when the chain is fixed 
















MECHANICS OF CARPENTRY 


317 


as in Fig. 12 as they are with the arrangement in Fig. 
13; or, the tighter the chain—i. e., the greater the angle 
between the forces B C and C A—the greater is the 
stress on the links. 

\ 

Example 5. A triangular bracket fixed against n 
wall, as shown in Fig. 16, has a weight of 5 cwts. sus¬ 
pended from the outer end o. What is the nature and 
amount of stress in each of the members o A and oB? 

Fig. 15 is the triangle of forces used to determine 
these stresses, and is drawn as follows: 1^ is drawn 



Fig. 15. 


Fig. 14. 


parallel to and represents the downward force (the 
weight of 5 cwts.) to scale. From 2 X draw 2^ parallel 
to 2 3 in Fig. 14 until it meets 1^ drawn parallel to 
1 3. Then the triangle l^^ represents the magnitude 
of the forces. 

By going round the triangle in order as shown by tl. 
arrows, we find that 2 3 acts towards the joint o an 
is therefore a compression stress or thrust, and 3 1 acts 
from the joint and is therefore a tension stress or pull. 











318 


MODERN CARPENTRY 


Fig. 16 shows a somewhat modified design of trian¬ 
gular wall-bracket, and Fig. 17 is the triangle of forces 
by which the stresses in the various members are ascer¬ 
tained. 



Fig. 17. 


Fig. 16. 


Example 6. What is the nature and amount of stress 
in each of the members AB and AC (Fig. 18) caused 
by the weight of 10 cwts. acting as shown? 

This example may be taken as typifying a simple 
kind of roof-truss with the weight taking the place of 
the ridge piece. Re-letter or figure the diagram accord¬ 
ing to Bow’s notation. Draw the vertical line 2yS 19 
equal in magnitude and direction to the weight 2 3. 
Complete the triangle by drawing lines parallel to the 
members A C and A B, from the points 2 X and 3 a re¬ 
spectively. These lines represent the amount of stress 
along the members A C and A B. On taking the sides 
of the triangle in order as shown by the arrows, it is 







MECHANICS OF CARPENTRY 


319 


seen that 2 1 3 1 act downwards; 3-Yi acts towards the 
joint A, as does also 1 1 2 1 ; therefore each member is 
subject to a compression stress (thrust). 

Fig. 19 shows another example of this kind with a 
much smaller angle between the forces. 

Fig. 20 illustrates a still further example, where the 
two sides are of unequal inclination. 



Polygon of Forces. The method of obtaining the 
resultant of any two forces acting at a point can be 
extended to three, four, or any number of forces. 

Example. 0 A, 0 B, 0 C, 0 D, 0 E, (Fig. 21) repre¬ 
sent the magnitude and direction of five forces acting 










320 


MODERN CARPENTRY 


at the point 0. Determine the magnitude and direction 
of the resultant force. 

This problem can be solved either by an application 
of the parallelogram of forces or by a direct construc¬ 
tion. 

(1) Determine by the parallelogram of forces, the 
resultant 0 1 of forces OA and OB (Fig. 22). Simi¬ 
larly, determine the resultant 0 2 of the forces 0 1 and 



0 C. Again, 0 3 is the resultant of the forces 0 2 and 0 
D ; and finally O 4 is the resultant of 0 3 and 0 E. There¬ 
fore, 0 4 is the resultant of all the original forces; or, 
in other words, a single force equal in magnitude and 
direction to the force 0 4 will have the same effect at 
the point 0 as the five forces have when acting to- 







MECHANICS OP CARPENTRY 


321 


gether. Since a force 4 0 will balance 0 4, a force rep¬ 
resented in magnitude and direction by the line 4 0 
will, together with the five given forces, produce equfi- 
ibrium at the point 0. 

(2) The same result may be obtained more simply as 
follows. Re-letter the forces as shown in italics (Fig. 
22), and then, as in Fig. 23, draw a straight line a'b' 



equal in magnitude and parallel to a b (Fig. 22). From 
b' draw T b'c' equal and parallel to b c; continue the pro¬ 
cess, taking the forces in order. It will be found by 
drawing the closing line of the polygon, that is, by 
joining x' to a', that x'a' gives the magnitude and the 
direction of the force required to produce equilibrium. 
Conversely, a'x' is the resultant of all the original 
forces. By drawing the line 4 0 through the point O 
(Fig. 22) and indexing it to scale, the required result- 




MODERN CARPENTRY 


322 


ant—which corresponds with the one determined by 
the parallelogram of forces—is obtained. Its direction 
is indicated by the arrow. 

Fig. 25 is the polygon of forces when two of the 
forces, be and d e, act towards the joint (Fig. 24), 
the magnitude of all the forces being as in the previous 
example. In this case the equilibrant is determined, 
and is shown by the thick line in Figs. 24 and 25. 



The polygon of forces may be stated as follows: If 

two or more forces act at a point, then, if starting at 
any point a line be drawn to represent the magnitude 
and direction of the first force, and from the point thus 
obtained another line be drawn similarly to represent 
the second force, and so on until lines have been drawn 
representing each force,—the resultant of all these 
forces will be represented by a straight line drawn 
from the starting point to the point finally reached. 

Polygons, parallelograms, or triangles, of forces, 
when used to determine either the resultant or the 
equilibrant of stresses acting at a point, are called 
reciprocal diagrams. 




MECHANICS OF CARPENTRY 


323 


Inclined Forces in one Plane but not acting* through 
one Point. The foregoing examples deal only with 
forces which act at a single given point, and in these 
cases the resultant acts at the same point. When all 
the forces do not act at the same point, the magnitude 
and direction of the resultant is obained as in previous 
examples, i. e., by drawing the reciprocal diagram; the 
line of action, however, still remains to be determined. 
To determine this line of action, it is necessary to draw 
what is known as the funicular or link polygon. The 
method is as follows: 



CL 


Fig. 27, 


Fig. 26 


Example. Let N, Y, Z (Fig. 26) be three forces in 
the same plane and of the magnitude and direction 
shown. It is required to find the magnitude and the 
line of action of the resultant force. 

Re-letter the forces abed according to Bow’s nota¬ 
tion and draw the reciprocal diagram a'b'c'd'; the line 





324 MODERN CARPENTRY 

a'd' which closes the figure represents the magnitude 
of the resultant. To obtain the actual line of action 
of the resultant, take any point or pole 0 and join 
a'O, VO, c'O, d'O. The figure thus obtained is called 
the polar diagram. The funicular polygon is now con¬ 
structed by drawing—anywhere in the space b—a line 
1 2 parallel to b'O and intersecting the forces X and Y 
at 1 and 2 respectively. From 2 draw 2 3 parallel to 
c'O, intersecting the force Z in 3. Through 1 draw 
1 4 parallel to a'O, and through 3 draw 3 4 parallel to 
d'O. Through the point of intersection 4, draw T a line 
R parallel to a'd'. R is the required line of action of 
the resultant of the three given forces, and its magni¬ 
tude is represented by the length of a'd'. 



Fig. 28. 


Parallel Forces. In addition to forces acting in the 
ways already explained, it is necessary to consider a 
few examples of parallel forces. (These must not be 
mistaken for those dealt with by the parallelogram of 
forces, as they are entirely different.) 

In all the examples of parallel forces now to be con¬ 
sidered, the forces will act vertically. As these can be 
shown easily both graphically and arithmetically, each 
example will be worked out by both methods. 

The simplest examples of the equilibrium of parallel 
forces are found in the use of levers. The lever shown 
in Fig. 28 is a straight bar resting on a triangular block, 






MECHANICS OF CARPENTRY 


325 


F, called a fulcrum. At the ends A and B of the lever, 
forces P and W respectively act vertically downwards. 
It is plain that forces P and W will tend to rotate the 
lever in opposite directions around the fixed point F. 
The tendency of either force to rotate the lever is called 
the moment of that force; it is measured by the product 
of the force into the perpendicular distance (called the 
arm of the force) of the fixed point from the line of 
action of the force. When the two moments are equal 
the lever is in equilibrium. The conditions of equil¬ 
ibrium therefore are: * 


PX A F=WXB F. 

If A F be 6", B F be 2", and W=9 lbs., then P will re- 

9X2 

quire to be —— = 3 lbs.: the moment of each force 
being 18 inch-lbs. 


A 

B 

M 

L 1 

L J 

6 

Zlbs 

0 

4lbs . 

Fig. 29. 



Since moments are always expressed in terms of the 
product of a force and a length, both these factors 
enter into every statement of the magnitude of a mo¬ 
ment. If the distances be expressed in feet, and the 
forces in cwts., the moments will, of course, be ex¬ 
pressed in ft.-cw T ts., and so on. 

Example 1 . A horizontal bar 3 ft. long has a weight 
of 2 lbs. at one end, and of 4 lbs. at the other end (Fig. 
29). Find the point at which the bar must be sup- 






MODERN CARPENTRY 


326 

ported so that it will rest horizontally. (Neglect the 
weight of the bar.) 

Arithmetically.—Since the lever is in equilibrium, 
the total downward force of 6 lbs. is balanced by an 
upward force (reaction) of 6 lbs. at the unknown point 
of support, and the moment of the upward reaction, 
about any point, is equal to the sum of the moments 
of the downward forces about the same point. Con¬ 
sider moments about A: 

Moment of weight at A, about A,=2X0. 

“ “ “ B “ A,=4XAB. 

reaction about A=(2-f4)XAX. 

6XAX=(2X0) + (4XAB); 

6AX=0-f- (4X3); 

AX=12/6=2 feet 

Graphically.—In the consideration of these forces 
graphically, the polygon of forces becomes a straight 
line A polar diagram and a funicular polygon are re¬ 
quired. 

Fig. 30 shows the bars with the weights suspended. 
Letter the forces A B and B C. The polar diagram is 
drawn as follows: Draw to a suitable scale, a vertical 
line a c, 6 units long—equal to the sum of the weights. 
From any point O which may be at any convenient dis¬ 
tance from a c, draw 0 a, Ob, and 0 c. To construct 
the funicular polygon, draw, in the space B, 1 2 parallel 
to b o. From 1 draw 1 3 parallel to a 0, and from 2 
draw 2 3 parallel to 0 c, and produce it to intersect 
1 3 in 3. Then the vertical line drawn through the 
point 3 will give the position of the fulcrum. If the 
distances from this point to the points of application of 


MECHANICS OF CARPENTRY 


327 


the weights be measured, it will be found that they are 
in inverse proportion to the magnitudes of the 



weights, and that the weight on the right hand side of 
the fulcrum, multiplied by its arm of leverage, will be 
equal to the weight on the left hand side, multiplied by 
the arm of leverage on that side—the arms being one 
and two feet respectively. 

A _ B C l> 

Zlbs. 3 lbs 6 lbs . 4U)S. 

Fig. 31. 

Example 2. Four weights of 2, 3, 6, and 4 lbs. re¬ 
spectively hang on a bar as shown in Fig. 31. Deter¬ 
mine the point at which the bar must be supported to 
rest horizontally, the weight of the bar being neglected. 













328 


MODERN CARPENTRY 


Arithmetically.—Let the required point of support 
he denoted by the letter X. When the bar is in equil¬ 
ibrium, the sum of the moments about A of the down¬ 
ward forces, must be equal to the moment, about A, of 
the upward reaction at the point of support. 

:. the downward moments about A 
= (2X0) + (3X7)+.(6X11) + (4X15) 

— 0 + 21 + 66 + 60 =147. 

The moment about A of the upward reaction 
=Sum of all the weightsXAX 
=(2+3+6+4) X AX=15AX; 

15AX=147; 

AX—147/15=9 4/5 feet. 



Graphically.—Draw the vertical line of loads e a, 
representing to scale the sum of the weights as shown 
(Fig. 32). Construct the polar diagram by drawing 
from any point 0 the lines e 0, d 0, c 0, b 0, and a 0. 










MECHANICS OP CARPENTRY 


329 


To draw the funicular polygon, draw vertical lines un¬ 
der each weight, and—starting anywhere in the line of 
the first weight as at 1—draw in the space B a line 1 2 
parallel to 0 b; in the space C draw 2 3 parallel to 0 c; 
in the space D draw 3 4 parallel to 0 d. Through 4 draw 
4 5 parallel to e 0 and through 1 draw 1 5 parallel to 
O a. The vertical line drawn through the point 5, 
where these two lines meet, gives the position of the 
point of support. 

Although the application of the lever as a tool or 
machine is an everyday occurrence with the workman 
in such appliances as the turning bar of the bench-vice 
and sash-cramp, the screw-driver, brace, pincers, claw- 


W=10Cwts 

W. 6 ft 



Fig. 33. 



hammer, grindstone, treadle lathe, mortising-machine, 
etc., the detailed consideration of each of these cannot 
be entered into for want of space. The following 
examples involving the use of the crowbar will suffice 
further to illustrate the principles involved. 

Example 1. What force must be exerted at one end 
of a crowbar 6 ft. long, to raise a weight of 10 cwts. at 
the other end: the bar resting on a fulcrum 9" from 
the weight. (Neglect the weight of the crowbar.) 

Let AB (Fig. 33) be the bar 6 ft. long, and F the 
fulcrum at 9" from A. Consider moments in inch-lbs. 
about F. Let x be the required force. 





330 


MODERN CARPENTRY 


Moment of 10 cwts. about F—9X10X112 incli-lbs. 

Moment of x about F=BF'X X = (72—9)X X ='63X X 
inch-lbs. 

• x= 9Xl0Xll2 =160 lbs. 

63 

Example 2. A man weighing 110 lbs. is using a 
crowbar 5 ft. long. What must be the position of the 
fulcrum to enable him to balance a weight of 1260 lbs. 
at the other end ? 

Let AB be the length of the bar; F the position of 
the fulcrum; and x the length of the long arm in inches; 
then (60—x) is the length of the short arm in inches. 

Taking moments about F, 

140x=1260 (60—x); 

140x=75600—1260x ; 

140x-j-1260x=75600; 

1400x=75600; 

x= =54 in.=4' 6"=length of long arm. 

P~200lbs. 

A 


Example 3. A lever 7 ft. long is used as shown in 
Fig. 34. If a force of 200 lbs. is applied at P in the 
direction of the arrow, what weight, placed at a point 
12" from the fulcrum, can be raised? 

. Taking moments in ft.-lbs. about F, 

WX 1=200X7; 

W=1400 lbs. 









MECHANICS OF CARPENTRY 


331 


Loaded Beams. The determination of the proportion 
of the total weight carried by each support of a loaded 
beam—in other words, the upward reaction of each 
support which is necessary to maintain equilibrium— 
affords a good practical example of the theory of paral¬ 
lel forces. 

Example 1 . A beam rests upon supports placed 8 
feet apart. A weight of 12 lbs. is placed on the beam 
at a distance of 2 ft. from the right-hand support. What 
proportion of the weight is carried by each of the sup¬ 
ports, the weight of the beam being neglected? 



Fig. 35. 


j 

Arithmetically.—In this case (Fig. 35) the downward 
force (weight) of 12 lbs. must be balanced by upward 
forces (reactions) at the points of support, respectively 
equal to the pressure at these points, and together equal 
to 12 lbs.; and the moments of the upward forces about 
the point c must be equal. 

:. Reaction at AXAc=Reaction at BXBc, 
i.e. Reaction at A :Reaction at B: :Bc :Ac, 


or Reaction at A: 


Sum of reactions 
at A and B 


: :Bc :(Bc-}-Ac), 


i.e. Reaction at A:12 lbs.:: Be :AB. 

This may be expressed in general terms as follows: 


, ,, , Distance of that 

Pressure on one end . that.. load fl0m 

caused by any load load 0 ther end 


Length between 
supports. 






332 


MODERN CARPENTRY 


therefore Pressure on A : 12 lbs. :: cB . AB; 


Pressure on A — 


12 X cB 

AB 


12X2 

8 


= 3 lbs. 


Similarly, Pressure on B — 


12 X Ac_12 X 6 

AB 6 


= 9 lbs. 


Graphically.—Pig. 36 shows the beam and supports 
with the load in position. The polar diagram is drawn 
as follows: Draw a vertical line a b, representing the 



weight (12 lbs.) to a suitable scale. From any point O, 
which may be at any convenient distance from a b, 
draw the triangle 0 a b. Draw as in the figure a verti¬ 
cal line directly under the load, and one under each 
point of support, as Z, x, y. Letter the load A B, and 
















MECHANICS OF CARPENTRY 


333 


the space between the supports C. These letters can 
now be used to denote the reaction at each point of sup¬ 
port—i. e., the upward force required to maintain 
equilibrium—which is equal and opposite to the pres¬ 
sure exerted on each support by the load. Anywhere 
in x, as from 1, draw, in the space A, a line 1 2, parallel 
to a 0; from 2, in the space B, draw 2 3 parallel to 
b 0. Join 1 3, and through the pole 0, draw 0 c paral¬ 
lel to 3 1. Then a c (on the vertical line of loads a b) 
represents to scale the pressure on the left-hand sup¬ 
port, and c b to the same scale represents the pressure 
on the right-hand support. 


Hcwls 7 Cwts. 


ZCrrtS, 

Q 




Fig - . 37. 


As the reaction at each end is equal in magnitude 
and opposite in direction to the pressure, a c gives the 
amount of the reaction A C, and b c gives the amount 
of the reaction B C; and the sum of the reactions—both 
acting upwards—is equal to the total weight (12 lbs.). 

Example 2. A beam is loaded as shown in Fig. 37. 
Determine the reaction at each end, that is, the upward 
force required at each point of support to maintain 
equilibrium. 

Arithmetically: 



. Reaction at A due_wt. at CXCB_5 X 13 

to weight at C AB 16 








334 


MODERN CARPENTRY 


Similarly, 

Reaction at A due_ wt. at DX DB— 7X_9. 

to weight at D AB 16 

Also Reaction at A due_ wt. at E X EB — 2 X 2 ^ 

to weight at E AB 16 

The total reaction at A is equal to the sum of the 
partial reactions as shown above; or it may be obtained 
directly thus: 

Total reaction at A 

(wt. at CXCB) + (wt. at DXDB) + (wt. at EX EB) 

- Air 

= (5X13) + (7X9)+(2X_2) = lg =8M ^ 

16 16 

Similarly, 

Total reaction at B 

(wt. at C X CA) + (wt. at DX DA) + (wt. at EX EA) 

“ AB 

. = (5X3)+(7X7)+(2X14) = | =5 ^ ^ 

Graphically.—Construct the vertical line of loads, 
representing to scale the sum of the weights as shown 
in Fig. 38. Fix the pole 0, and draw the dotted lines 
0 a, 0 b, 0 c, 0 d. Letter the loads and draw a dotted 
vertical line directly under each load and under each 
support as shown. From any point in the first line, 
draw in the space A, a line 1 2 paralel to a 0; from 
2 draw 2 3 parallel to b 0; from 3 draw 3 4 parallel to 
c 0; and in the space D, from 4 draw 4 5 parallel to 
d 0. Join 1 to 5, thus completing the funicular poly¬ 
gon. By drawing a line parallel to 5 1—the closing 
line of this polygon—through pole 0, and meeting the 
vertical line of loads at e, it is found that e a equals 















MECHANICS OP CARPENTRY 


335 


the reaction E A, and e d equals the reaction E D; they 
are together equal to the sum of the weights in the 
beam. 



4a c J)^ £ f m 


6 Grt* 5 fats. 4 fats. 3cwts. 

OOP o 


ib 


Fig. 39. 


Example 3. A beam weighing 6 cwts. is loaded as 
shown in Fig. 39. Determine the reaction at each end 
necessary to produce equilibrium. 















336 MODERN CARPENTRY 

When the weight of a uniform beam is to be consid¬ 
ered, it may be taken as acting half-way between the 
supports, and thus adding half its weight to each sup- 

6cwts 5Cwis 4-Cwts. 3 Cwts 














MECHANICS OF CARPENTRY 


337 


Stress Diagrams for Roof Trusses.— Figs. 41 to 50 
show an application of the foregoing graphic methods 
to the determination of stresses in roof trusses. Fig. 
41 is a line diagram of a king-post truss loaded in the 
usual way. It must be noticed that the lettering is ar¬ 
ranged so that every member is indicated by a letter 
on each side. It is first necessary to determine the 
amount of weight carried by each point of support. 
This example is simplified by the symmetrical loading, 
as one half the weight is carried by each point of sup¬ 
port. When this is not the case, the proportion of the 
weight carried by each support must be determined 
first, by a consideration of parallel forces, as in earlier 
examples. 

It is usual when determining the stresses of such a 
truss, to draw the stress diagram, shown in Fig. 46. 
This diagram is a combination of Figs. 42 to 45, which 
are only drawn as separate figures to assist in under¬ 
standing the question more clearly. 

Fig. 42 is the polygon of forces for the joint (1) at 
the foot of the principal rafter on the left. Four forces 
act at the point: A B downwards, B N the principal 
rafter, N G the tie beam, and the upward force. A G— 
the reaction at the point of support. Of these four 
forces the amounts of two, AB and AG, are known; 
it is required to determine the nature and amounts of 
the stress of B N and of N G when acting at the angles 
given. 

Commencing Fig. 42 by drawing ab equal to A B, 
and a g equal to A G, the upward force. As these two 
forces are in the same straight line, and in opposite 
directions, their resultant is the line b g. From b 


338 MODERN CARPENTRY 








Fig. 44. 





















MECHANICS OF CARPENTRY 


339 


draw b n parallel to BN; and from g draw g n parallel 
to G N until b n and g n meet. Then b n g is the poly¬ 
gon (a triangle in this case) of forces acting at the 
point; and b n and n g represent the amount of stress 
in the principal rafter and tie-beam respectively. 

The direction of the stress is found by taking the 
forces in order: thus, g b acts upwards; b n acts to¬ 
wards the joint, therefore this member is in compres¬ 
sion; n g acts from the joint, which indicates that the 
tie-beam is in tension. 

At joint (2), four forces act, namely, B C, B N, C M, 
and N M. The two known forces are B C acting down¬ 
wards and B N towards the joint. For the magnitude 
of the stress on B N has already been found, and its 
direction of action at joint (2) is the opposite to its 
direction at joint (1). Fig. 43 shows the application 
of the polygon of forces to this joint. In it, b c and 
b n are drawn equal and parallel to B C and B N re¬ 
spectively; and, by drawing nm parallel to N M and 
c m parallel to C M, the stress diagram is obtained. 
This shows that the stress in C M, the upper part of the 
principal rafter, is much less than in B N, the lower 
part. By tracing the polygon, it is found that b c acts 
towards the joint, c m towards the joint (therefore C M 
is in compression), mn towards the joint (compression) 
and n b towards the joint (compression as in the pre¬ 
vious figure). 

At joint (3) there are four forces, i. e. C M, C D, 
D L, M L, acting as shown. Of these four forces the 
two C M and C D are known. Since the amount and 
nature of the stress in any member must be the same 
at any intermediate point between the joints, the stress 


340 


MODERN CARPENTRY 



Fig. 47. 




Fig. 49. 


Fig. 50. 






















MECHANICS OF CARPENTRY 


341 


in C M acting upon joint (3) must be as determined by 
the diagram for joint (2). Fig. 44 is the stress dia¬ 
gram ; e d is drawn parallel and equal to CD; cm 
parallel and equal to C M; m 1 and d 1 are drawn paral¬ 
lel to M L and L D respectively until they meet. Tak¬ 
ing these forces in order, c d is towards the joint, D L 
towards the joint (compression), L M is from the joint 
(tension), and MC towards the joint (compression). 

The tension stress in L M is caused by the struts 
M N and L H, which transfer part of the loads B C and 
D E respectively to the foot of the king-post. If no 
struts existed in this truss there would be no stress in 
ML. 

Joint (4) has five forces acting, each one of which 
has already been determined, since the stress diagrams 
for one side of the truss are in this example applicable 
to each side. For example, the diagrams showing the 
stresses in the joints (1) and (2) are applicable to 
(6) and (5) respectively. An examination of Fig. 45 
will show that g n is parallel and equal to GN; n m is 
parallel and equal to N M; 1 m is parallel and equal to 
LM; and lh being drawn parallel to LH meets mn 
in n, whilst h g is equal to n g. The diagram therefore 
shows the stress in each of the five members. 

In Fig. 46, which is the complete stress diagram for 
the members of the truss, the lettering is identical with 
that in each of the separate Figs. 42 to 45, and will be 
easily understood from them. 

Fig. 47 is a line diagram of a queen-post roof truss, 
and Fig. 48 is the stress diagram of this truss. Simi¬ 
larly, Figs. 49 and 50 are respectively the line diagram 
of and the stress diagram for, a composite roof truss, 


342 


MODERN CARPENTRY 


sometimes named a German truss. The detailed ex¬ 
planation already given will enable the figures to be 
understood. 


STRENGTH OF WOODEN BEAMS. 

For the purpose of calculating the carrying capacity 
of wooden beams, it is necessary to notice the nature of 
the stresses to which they are subjected, as well as the 
manner in which they are loaded, and the arrangement 
of the load. 

Stress and Strain. When a weight, or other force, 
acts upon a beam, it tends to change the shape and 
size of the beam. The force is technically called a 
stress, while the change in shape or size is called a 
strain. When a beam or girder, supported at both 



Fig. 51. 


ends, is loaded, the upper part tends to shorten. The 
lower fibres, on the other hand, are in a state of ten¬ 
sion, as they tend to stretch. The force acting on the 
upper fibres of such a beam is therefore a compression 
stress; that on the lower fibres is a tension stress. 

The existence of these stresses may be made very 
apparent either by making a saw-cut across, or by 
actually cutting out a wedge-shaped piece from, the 
middle of a beam of wood for half its depth, as shown 
in Fig. 51. On resting the beam on two supports with 
the cut edge uppermost, and then loading it, it will be 











MECHANICS OF CARPENTRY 


343 


seen that the saw-cut closes. This shows that the fibres 
on the upper side are in a state of compression. If the 
same beam is now turned over so that the saw-cut is on 
the lower side, and again loaded, the tendency is for 
the cut to open, thus showing that the fibres on the 
lower side are in a state of tension. 



Shearing* Stresses. A shearing stress is one which 
gives the fibres of the wood a tendency to slide over 
one another. A shearing stress may be either in the 
direction of the fibres or at right angles to them. To 
illustrate a shearing stress in the direction of the fibres 
of a beam, imagine the beam cut into a number of 








1 


< 



I 



Fig. 53. 


boards; place these on the top of each other in the 
position of a beam resting upon supports at each end, 
and place a load in the middle. The result will be that 
the beam will bend as shown in Fig. 52, and the boards 
will slide over each other. A shearing stress across 
the fibres of a loaded beam can be illustrated by tak¬ 
ing a bar of soap, or some such soft material, resting 
it upon supports, and loading it. The result will be as 
shown in Fig. 53. 




















344 


MODERN CARPENTRY 


Methods of Arranging Beams. The nature and 
amount of stress in the fibres of a loaded beam depend 
upon the way the beam is supported and on the ar¬ 
rangement of the load. Thus a cantilever is a beam 
with one end only secured upon a support, the other 
end overhanging. The load upon a cantilever may be 
a concentrated load at the outer end, as in Pig 57, or 
the load may be anywhere between the outer end and 


: 1 

1 

Tension 

1 

mw'v 

1 $ 

• 

• 1 

Compress id re -- J 


Fig. 54. 


the supported end; a number of loads of varying 
weights may be distributed over the length; the load 
may be a uniformly distributed one extending over the 
length of the beam, or it may be a combination of a 
concentrated load and a distributed load. A cantilever 
loaded in any of the ways just described has the fibres 
in the upper edge in a state of tension, those in the 
lower half being in compression (Fig. 54). 


Compressions 



■ - 1 n -— 


•asm 


Tvjtsww - “ 



Fig. 55. 

A beam supported at both ends may be loaded in any 
of the ways described for the cantilever, with the re¬ 
sult that the stresses will be as shown in Pig. 55 ; i. e. 
the upper part will be in compression, and the lower 













MECHANICS OF CARPENTRY 


345 


half in a state of tension. The stresses in the various 
parts of a loaded beam which has the ends fixed differ 
from those of the beam which simply rests upon sup¬ 
ports. They are illustrated in Fig. 56, which shows 
that for a distance of about one-fourtli from each end 
the beam takes the form of a cantilever, and has the 
fibres in the upper half in a state of tension and the 
lower fibres in compression. The remainder of the 
beam has the upper fibres in compression and the lower 
part in tension. The neutral axis of all these beams 
is in the centre of the depth. If a long beam has inter¬ 
mediate supports as in Fig. 56, it may be regarded as 
being “fixed” at the points of intermediate support. 


4--T- ‘--Hrr- C- C- ye- T - 

r r-r - " 



I 

C- ^ V* — £*-•"> 

1 

-- - -’jj 


1 

5*^ 

1 



Fig. 56. 


Bending Moments. For the purpose of making com¬ 
parisons of the relative strengths of loaded beams, a 
further consideration of the “moment of a force” is 
necessary. Since the tendency to bending, to which a 
given beam is subject at any point, depends upon the 
moments of the stresses about that point, it is obvious 
that the relative strengths of beams may be measured 
in terms of moments. The bending moment at any given 
section is the algebraic sum of all the external forces 
acting on one side of the section. Since it is at the 
point where the greatest bending moment occurs that 
the beam is subjected to the greatest stress, it follows 
that it is of some importance to be able to determine 








346 


MODERN CARPENTRY 


the bending moment of beams loaded under different 
conditions. The bending moment—like other moments 
—must always be expressed in terms of a length and 
a force. 

Example 1 . A cantilever carries a load of 6 tons at 
its outer end, which is 5 ft. from the supporting wall. 
Determine the maximum bending moment, and also the 
bending moment at 2 ft. from the wall. 



The greatest tendency to bending will be at the point 
of support, i. e. at a distance of 5 ft. from the load. 
Maximum bending moment=5X6=:30 ft.-tons. 
Bending moment at 2 ft. from the wall (i. e. 3 ft. 
from the load) =3X6=18 ft.-tons. 

The bending moment at any distance from the load 

* 

may be determined graphically by drawing, as in Fig. 
57, a vertical line A B 30 units long (respecting the 
maximum bending moment) under the point of support 
A (i. e. the point where the bending moment is a maxi¬ 
mum), and joining B C. Then the bending moment at 
the point a will be represented by the length of a b 
drawn parallel to A B. 








MECHANICS OF CARPENTRY 


347 


Example 2. A cantilever projects 4 ft. and carries a 
uniformly distributed load of 8 cwts. along the upper 
edge. Determine the maximum bending moment, and 
also draw a diagram from which the bending moment 
at any section along the length of the cantilever may 
be determined. 



Fig. 58. 


A load arranged as shown in Fig. 58 is equivalent 
to a concentrated load of 8 cwts. acting in the middle 
of the length, i. e. 2 ft. from the point of support. The 
maximum bending moment will therefore be 

8X2=16 ft.-cwts. 

Fig. 58 is the diagram from which the bending mo¬ 
ment at any section may be determined. The load is 
supposed divided into 4 equal parts, and the bending 
moment due to each part is drawn to scale on the verti- 
















348 


MODERN CARPENTRY 


cal line A E. The weight of Z acts at 3' 6" from A, 
and the maximum bending moment due to Z=2X3.5= 
7 ft.-cwts. Draw A B 7 units long. Similarly, the 
maximum bending moment due to Y=2X2.5=5 ft.- 
cwts., and is represented by B C ; maximum bending mo¬ 
ment due to X=2Xl-5=3 ft.-cwts., represented by C D ; 
maximum bending moment due to W=2X0.5=1 ft.- 
cwt., represented by D E. The maximum bending mo- 




Fig. 60. 




Fig. 63. 


Fig. 62. 


ment due to the total load is therefore 7+5+3+1 = 16 
ft.-cwts., and is represented by A E. Draw a vertical 
line through the centre of each part of the load, and 
complete the triangles A a B, B b C, C c D, D d E. 
Draw an even curve touching the lines E d, D c, C b, 
B a. This curve is a parabola. The bending moment 




























MECHANICS OP CARPENTRY 349 


at any section P is represented by the length of the 
vertical line P Q cutting the parabola at Q. 

The following formula} are used for determining the 
relative bending moments (and therefore the relative 
strengths required) of beams loaded in various ways. 
In each case L=the length of-the beam and W=the 
weight of the load. 


Cantilever fixed at one end and loaded ) 
at the other end (Fig. 57), j 

Cantilever fixed at one end and loaded } 
with uniformly distributed load, \ 


Maximum 

Bending 

Moment. 


Relative 

Strength. 


WL 1 


WL 

2 



Beam supported at both ends and j WL 

loaded with a central load (Fig. 59), J 4 


Beam supported at both ends and 
loaded with a uniformaly distributed r 
load (Fig. 65), J 

Beam fixed at both ends and loaded I 
with a central load (Fig. 62), ) 

Beam fixed at both ends and loaded \ 
with a uniformily distributed load, J 


WL 

8 

WL 

8 

WL 

12 


8 

8 

12 


Figs. 60 to 63 show beams loaded in various ways, 
and serve to illustrate the method of determining 
graphically the bending moment at any section of the 


beam. 


It will be noticed that the maximum bending moment 
of a beam supported at each end is in each case in the 
line of the load, and with a central or an evenly dis¬ 
tributed load is at the middle of the length of the beam. 

Calculation of the Transverse Strength of Wooden 

Beams. Other things being equal, the strength of a 







350 


MODERN CARPENTRY 


rectangular wooden beam is directly proportional to 
the breadth in inches multiplied by the square of the 
depth in inches, and inversely proportional to the 
length in feet. Of course the nature of the material 
is also an important factor, since timber, even of the 
same kind, varies in strength to a considerable extent. 
Each beam therefore has what is called a natural con¬ 
stant, which must be considered in the calculation of 
its carrying capacity. To obtain this constant, it is 
usual to take a bar of similar wood, 1 inch square in 
section, and long enough to allow of its being placed 
on supports 1 foot apart. The constant is the weight 
of the central load, which is just sufficient to break 
the bar. The constant may be expressed in lbs., cwts., 
tons, etc, and the carrying capacity will always be in 
the same terms. The following constants (in cwts.) 
may be adopted for the purposes of calculation: oak, 
ash and pitch pine, 5; red deal, red pine and beech, 4; 
white and yellow pine, 3. 

Another important consideration is the ratio which 
the breaking load of a beam bears to the “safe” load. 
This ratio is called the factor of safety, and its value 
depends upon whether the load is a live—a constantly 
moving—load, or a dead (i. e., a stationary) load. The 
factor of safety for a dead load is usually taken at 5, 
which means that the safe load upon a beam must not 
exceed one-fifth of the breaking load; the factor of 
safety for a live load is often taken at 10. 

For beams supported at both ends the formula 

bd 2 c 

W = —— may be used for the purposes of calcula- 
E 

tion when: 



MECHANICS OF CARPENTRY 


351 


W=breaking weight or maximum carrying capacity 
of a centrally loaded beam, expressed in the 
same terms as the constant, 
b=breadth of the beam in inches, 
cl=depth of the beam in inches, 

L=length of the beam in feet, 

c=the constant, found by experiment as described 
above, and expressed in terms of lbs. or cwts. 
To illustrate the above formula, take two pieces of 
the same kind of wood say 7 ft. long, 6 in. wide and 2 
in. thick. Place one of these pieces flat, and the other 
on one edge, the distance between the supports in each 
case being 6 ft. As the constant is the same in both 
(say 5 cwts.), the carrying capacity of each will be 

bd 2 c. 

expressed by the formula W= ~j~ > 


6 X 2 X 2 X5 ^ 
for the flat beam, W — g — 20, 

2X6X6X5__ 

for the one on edge, W — x — 60 


and the relative strengths will be as 20: 60 or as 1:3. 
When it is necessary to find other terms than W, 


the equation W = 


bd 2 c 

L 


may be expressed as follows: 


bd 2 c , WL J9 _ WL 3 _ IWL _ WL 

L _ W ’ b d 2 c ’ C be ’ f \ be ’ C bd 2 

The value of W for a distributed load is twice that 


2 bd*c __ 

for a concentrated load, i. e., W— • When the 

ends are fixed the carrying capacity is increased by 
about one-half. 













352 


MODERN CARPENTRY 


bd 2 c 

For safe central loads the formula W= is used: 

F being the factor of safety. 

Example 1. Find the maximum carrying capacity 

of a centrally loaded wooden beam of pitch pine, 13 

ft. long (12. ft. between the supports), 10 in. wide, 

and 6 in. thick, (1) when placed on edge; (2) when 

placed flat. Assume a constant of 5 cwts. 

Applying the formula 

bd 2 c , x „ T 6 X 10 X 10 X 5 OCA , . 

W=—j—, (1) W =- no -=250 cwts.when on edge. 

L 


(2) W= 


12 

10 X 6 X 6X5 


12 


“ 150 cwts. w T hen placed flat. 


Example 2. What would be the maximum safe load 
to which the beam in Ex. 1 may be subjected (1) as a 
central load; (2) as a uniformly distributed load? 

Formula for safe central load using a factor of safety 
of 5, is 

bd 2 c 


Safe central load =- 


LF 


6 X 10X 10X 5 


12X5 
10 X 6 X 6 X 5 


= 50 cwts. for beam on edge, 


= 30 cwts. for beam placed flat. 


or 12 X 5 

Formula for safe uniformly distributed load, again 


using factor of safety of 5, is 


Safe distributed load = 


2bd 2 c 


LF 


2X6X10X10X5 


12X5 
2 X 10 X 6 X 6X5 


— 100 cwts. for beam on edge, 


or 


12X5 


= 60 cwts. for beam placed fl«t. 













MECHANICS OF CARPENTRY 


O'JO 


Example 3. Find the breadth of a beam of oak rest¬ 
ing upon supports 18 feet apart, the beam being 12 in. 
deep, to carry safely a uniformly distributed load of 
5 tons. Constant 5 cwts. 


Safe distributed load, W = 


2bd 2 c 


LF 


, _WLF_(5 X 20)X 18 X 5_25_ _ , 

‘ * b 2d 2 c 2 X 12 X 12 X 5 4 6/ * mches - 

Example 4. A beam of red or yellow pine 20 ft. 
long (between supports), and 10 in. broad has to carry 
safely (1) as a central load, (2) a distributed load of 4 
tons. What must be the minimum depth of the beam 
in each case? (Constant 4 cwts.) 

WLF. 


With a central load d 2 =- 


bc 


, /WLF_ / 

d= \ _ bc \ 


With a distributed load 


80 X 20 X 5 


10X4 


=1/200=14.14 inches. 




80 X 20 X 5 
2X10X4 


=i/100=10 inches. 


Example 5. —What size of beam is required to carry 
safely a central load of 35 cwts. over a 10 ft. span; 
the depth and breadth of the beam being in the pro¬ 
portion of 7 : 5? (Constant 5 cwts.) 


b=fd; 
WLF 


d 2 b= 


. , 2 5 WLF 

i.e. d 2 .^~d—- 


(7 
















354 


MODERN CARPENTRY 


5cP_WLF 
7 c 

7VVLF 7 X 35 X 10 X 5_ 
d -5X5 4 °°- 

d=V490=7.88"=nearly 8". 

, 5X7.88 r „. u 

b=- -j -=5.6 inches. 

The strength of flitched girders may be calculated 
by considering the wooden beam and iron flitch sepa¬ 
rately. The thickness of the flitch is usually about 
1-12 that of the wooden beam. The constant for 
wrought iron is 25 cwts. 

Deflection. In arranging beams it is necessary to 
consider not merely the strength of the beam, but also 
its liability or otherwise to be bent out of shape—or 
deflected—by the load placed upon it; since a beam 
which is overloaded and bent to a large extent has the 
fibres strained and therefore permanently weakened. 
The resistance which a beam offers to deflection is 
called its stiffness. It should be noticed that the 
“strongest” beam is not necessarily the “stiffest,” nor 
the stiffest beam the strongest. 

It is of importance to be able to determine the cross- 
sections of the strongest and the stiffest beams re¬ 
spectively which can be cut from a given log. 

Suppose the log is of circular cross-section. 

(a) To find the cross-section of the strongest beam. 
—Draw a diameter AB (Fig. 64) and divide it into 3 
equal parts at 1 and 2. Fbom 1 and 2 draw perpen¬ 
diculars to AB cutting the circumference at C and D 
respectively. Join ACBD. The rectangle ACBD is the 
section required. 


JO 






MECHANICS OF CARPENTRY 


355 


(b) To find the cross-section of the stiffest beam.— 
Divide the diameter AB (Fig. 65) into 4 equal parts 
at 1, 2, 3, and draw 1C and 3D perpendicular to AB, 
and cutting the circumference in C and D respectively. 
The rectangle ACBD is the section required. 

Since the strength of a beam is proportioned to 


bd 2 
L ’ 


and the value of this fraction increases as d 


increases when bd (i. e., the sectional area) remains 



Fig. 64. 


Fig. 65. 


constant, the strongest beam of any given sectional 
area would be that of greatest depth if the tendency 
to buckling could be avoided. In the case of floor 
joists the ratio of depth to breadth is often as 3:1 
or even 4:1, and the tendency to buckling is over¬ 
come by strutting. The strongest beam is that which 
has the depth to the breadth as 7:5. 






356 


MODERN CARPENTRY 


Pulleys. It is necessary to consider one or two sim¬ 
ple arrangements by which pulleys are used for hoist¬ 
ing purposes. In the following examples the fiiction 
will for the sake of simplicity be neglected, although 
in practice it must be taken into account. Fig. 66 
illustrates the simplest application of the pulley. It is 



Fig. 66. 



plain that when the forces acting on the pulley are in 
equilibrium they are equal, and the only advantage 
gained is in the change of direction of the force re¬ 
quired to balance W. Therefore in this example P—W. 

In Fig. 67 the force balancing P is the tension of 
the cord A, which is equal to that of B. The sum of 




































MECHANICS OF CARPENTRY 


357 


these two equal tensions is plainly equal to the weight 


W. Therefore p = —, 

2 ’ 

is 2. 


and the mechanical advantage 


Figs. 68 and 69 are illustrations of a two- and three- 
sheaved pulley block respectively. By arranging pul¬ 
leys side by side in this manner, and using a combina¬ 
tion of two similar blocks as in Fig. 70 a mechanical 
advantage equal to the number of pulleys around 
which the rope passes is obtained. In other words, the 




Fig. 68. 


Fig. 69. 


power required is equal to the weight raised divided 
by the number of pulleys around which the rope 
passes. Thus with 3 pulleys in each block there 
will be six cords, and the power required to bal¬ 
ance a weight of 18 cwts. will be 18 -f- 6 = o cwts., 
plus the force required to overcome friction. 















358 


MODERN CARPENTRY 


Specific Gravity. The specific 
gravity, or relative density, of a 
body is the ratio of the weight of 
that body to the weight of an equal 
volume of water. Thus a block of 
wood weighing 40 lbs. per cubic foot 

40 

has a specific gravity of —— = 0.64 

62.5 

(since a cubic foot of water weighs 
62.5 lbs.). 

When a body floats in water, and 
is therefore in equilibrium, the 
weight of the body is balanced by 
an equal upward reaction, the 
weight of the water displaced being 
equal to the total weight of the 
floating body. 

Example. A block of wood 9"x9" 
x9", floats in water with its upper 
surface 2.5" above the surface of the 
water. Find its specific gravity. 

of the block is submerged. 

9 lo 

By definition, the specific gravity 
of the wood is the ratio of the 
weight of any portion of the block 
to the weight of an equal volume of water. Consider 
the part of the block below the surface of the water. 

_Weig ht of submerged part of wood 

Weight of displaced water 

_ Wei ght of submerged part of wood 

Weight of whole block 


W 


Fig. 70. 


Specific gravity 






















MECHANICS OF CARPENTRY 359 


__ Volume of submerged part of wood 

Volume of whole block' 

- 1 = 

In addition to the foregoing I have thought it es¬ 
sential to add the following special notes which are 
the results of actual experiments, and which will, I 
am sure, appeal to the American workmen, especially 
those who are located in the West, where timber is 
yet often employed in heavy structures. 


STRENGTH OF TIMBER. 

Proposition I. The strengths of the different pieces 
of timber, each of the same length and thickness, are 
in proportion to the square of the depth; but if the 
thickness and depth are both to be considered, then 
the strength wull be in proportion to the square of the 
depth, multiplied into the thickness; and if all the 
three dimensions are taken jointly, then the weights 
that will break each will be in proportion to the square 
of the depth multiplied into the thickness, and divided 
by the length; this is proved by the doctrine of 
mechanics. Hence a true rule will appear for pro¬ 
portioning the strength of timbers to one another. 

Rule. Multiply the square of the depth of each piece 
of timber into the thickness; and each product being 
divided by the respective lengths, will give the pro¬ 
portional strength of each. 

Example. Suppose three pieces of timber, of the 
following dimensions: 

The first, 6 inches deep, 3 inches thick, and 12 feet 
long. 




360 MODERN CARPENTRY 

The second, 5 inches deep, 4 inches thick, and 8 feet 
long. 

The third, 9 inches deep, 8 inches thick, and 15 feet 
long. The comparative weight that will break each 
piece is required. 

Ans. 9, 12Y2, and 43 1-5. 

Therefore the weights that will break each are 
nearly in proportion to the numbers 9, 12, and 43, 
leaving out the fractions, in which you will observe, 
that the number 43 is almost 5 times the number 9; 
therefore the third piece of timber will almost bear 5 
times as much weight as the first, and the second piece 
nearly once and a third the weight of the first piece; 
because the number 12 is once and a third greater 
than the number 9. 

The timber is supposed to be everywhere of the same 
texture, otherwise these calculations cannot hold true. 

Proposition II. Give the length, breadth, and 
depth of a piece of timber; to find the depth of 
another piece whose length and breadth are given, so 
that it shall bear the same weight as the first piece, 
or any number of times more. 

Rule. Multiply the square of the depth of the first 
piece into its breadth, and divide that product by its 
length: multiply the quotient by the number of times 
as you would have the other piece to carry more weight 
than the first, and multiply that by the length of the 
last piece, and divide it by its width; out of this last 
quotient extract the square root, which is the depth 
required. 

Example I. Suppose a piece of timber 12 feet long, 
6 inches deep, 4 inches thick; another piece 20 feet 


MECHANICS OF CARPENTRY 


361 


long, 5 inches thick; required its depth, so that it shall 
bear twice the weight of the first piece. 

Ans. 9.8 inches, nearly. 

Example II. Suppose a piece of timber 14 feet long, 
8 inches deep, 3 inches thick; required the depth of 
another piece 18 feet long, 4 inches thick, so that the 
last piece shall bear five times as much weight as the 
first. 

Ans. 17.5 inches, nearly. 

Note.—As the length of both pieces of timber is 
divisible by the number 2, therefore half the length 
of each is used instead of the whole; the answer will 
be the same. 

Proposition III, Given the length, breadth, and 
depth, of a piece of timber; to find the breadth of 
another piece whose length and depth is given, so that 
the last piece shall bear the same weight as the first 
piece, or any number of times more. 

Rule. Multiply the square of the depth of the first 
piece into its thickness; that divided by its length, 
multiply the quotient by the number of times as you 
would have the last piece bear more than the first; 
that being multiplied by the length of the last piece, 
and divided by the square of its depth, this quotient 
will be the breadth required. 

Example I. Given a piece of timber 12 feet long, 
6 inches deep, 4 inches thick; and another piece 16 feet 
long, 8 inches deep; required the thickness, so that it 
shall bear twice as much weight as the first piece. 

Ans. 6 inches. 

Example II. Given a piece of timber 12 feet long, 
5 inches deep, 3 inches thick; and another piece 14 


362 


MODERN CARPENTRY 


feet long, 6 inches deep; required the thickness, so* 
that the last piece may bear four times as much weight 
as the first piece. 

Ans. 9.722 inches. 

Proposition IV. If the stress does not lie in the 
middle of the timber, but nearer to one end than the 
other, the strength in the middle will be to the strength 
in any other part of the timber, as 1 divided by the 
square of half the length is to 1 divided by the rect¬ 
angle of the two segments, which are parted by the 
weight. 

Example I. Suppose a piece of timber 20 feet long, 
the depth and width is immaterial; suppose the stress 
or weight to lie five feet distant from one of its ends, 
consequently from the other end 15 feet, then the above 

portion will be ^^10 = I5o : 5^15 = A as the 
strength at five feet from the end is to the strength 


at the middle, or ten feet, or as 


100 _ . 100 _ 1 
100 — 1 ■ 75 ~ 1 3' 


Hence it appears that a piece of timber 20 feet long 
is one-third stronger at 5 feet distance for the bearing, 
than it is in the middle, which is 10 feet, when cut in 
the above proportion. 

Example II. Suppose a piece of timber 30 feet long; 
let the weight be applied 4 feet distant from one end, 
or more properly from the place where it takes its 
bearing, then from the other end it will be 26 feet, 


and the middle is 15 feet; then, 


1 


1 


1 


15X15 225 4x26 


225 1 225 0 17 , Q 1 

or as -= 1 : — = 2-or nearly 2-. 

104 225 104 104 J 6 













MECHANICS OF CARPENTRY 


363 


Hence it appears that a piece of timber 30 feet long 
will bear double the weight, and one-sixth more, at 
four feet distance from one end, than it will do in the 
middle, which is 15 feet distant. 

Example III. Allowing that 266 pounds will break 
a beam 26 inches long, requireth the weight that will 
break the same beam when it lies at 5 inches from 
either end; then the distance to the other end is 21 
inches; 21X5=105, the half of 26 inches is 3:.13X13 
=169; therefore the strength at the middle of the 
piece is to the strength at 5 inches from the end, as 


169 . 169 
169 ' 105 


or as 1 : 


169 

105 


: : 266 : 428 + 


169 

105 


the proportion is stated thus: 


1 : 


Ans. 


. From this calculation it appears, that rather more 
than 428 pounds will break the beam at 5 inches dis¬ 
tance from one of its ends, if 266 pounds will break 
the same beam in the middle. 

By similar propositions the scantlings of any timber 
may be computed, so that they shall sustain any given 
weight; for if the weight one piece will sustain be 
known, with its dimensions, the weight that another 
piece will sustain, of any given dimensions, may also 
be computed. The reader must observe, that although 
the foregoing rules are mathematically true, yet it is 
impossible to account for knots, cross-grained wood, 
&c., such pieces being not so strong as those which are 
straight in the grain; and if care is not taken in 
choosing the timber for a building, so that the grain 
of the timbers run nearly equal to one another, all 
rules which can be laid down will be baffled, and con- 





364 


MODERN CARPENTRY 


sequently all rules for just proportion will be useless 
in respect to its strength. It will be impossible, how¬ 
ever, to estimate the strength of timber fit for any 
building, or to have any true knowledge of its propor¬ 
tions, without some rule; as without a rule everything 
must be done by mere conjecture. 

Timber is much weakened by its own weight, except 
it stands perpendicular, which will be shown in the 
following problems; if a mortice is to be cut in the 
side of a piece of timber, it will be much less weak¬ 
ened when cut near the top, than it will be if cut at 
the bottom, provided the tenon is driven hard in to 
fill up the mortice. 

The bending of timber will be nearly in proportion 
to the weight that is laid on it; no beam ought to be 
trusted for any long time with above one-third or one- 
fourth part of the weight it will absolutely carry: 
for experiment proves, that a far less weight will 
break a piece of timber when hung to it for any con¬ 
siderable time, than what is sufficient to break it when 
first applied. 

Problem I. Having the length and weight of a beam 
that can just support a given weight, to find the length 
of another beam of the same scantling that shall just 
break with its own weight. 

Let l=the length of the first beam, 

L=the length of the second; 
a^=the weight of the first beam, 
w=the additional weight that will break it. 

And because the weights that will break beams of the 
same scantling are reciprocally as their lengths, 


MECHANICS OF CARPENTRY 


365 


therefore — 




w -f- 

“l 


n 

2 

-1 = W = the weight that 


a 


w T ill break the greater beam; because w + ris the whole 

A 


weight that will break the lesser beam. 

But the weights of beams of the same scantling are 
to one another as their lengths: 


Whence,! : L : : | = W half the weight of the 


greater beam. 

Now the beam cannot break by its own weight, 
unless the weight of the beam be equal to the weight 
that will break it: 


Wherefore, 


, A 

La W _2 _ 2w+a. 

21 L 2L 1 


L 2 a = 2wl 2 + al 2 , 


a: 2 w+a:: l 2 : L 2 , consequently VL 2 =L=the length 
of the beam that can just sustain its own weight. 

Problem II. Having the weight of a beam that can 
just support a given weight in the middle, to find the 
depht of another beam similar to the former, so that 
it shall just support its own weight. 

Let d=the depth of the first beam; 
x=the depth of the second; 
a=the weight of the first beam; 
w=the additional weight that will break the first 
beam; 


then will w+^ or a - = the whole weight that will 

A A 


break the lesser beam. 









366 


MODERN CARPENTRY 


And because the weights that will break similar beams 
are as the squares of their lengths, 



2w + a 
2 


2x 2 X w + ax 2 

“2d 2 


= W 


the weights of similar beams are as the cubes of their 
corresponding sides: 

Hence d 3 : x 3 : : ~ : 


. ax 3 2x 2 w + x 2 a 
' , 2d 3= 2d 2 


.’. ax = 2wd + ad 


a: a-j-2 w:: d: x=the depth required. 

As the weight of the lesser beam is to the weight of 
the lesser beam together with the additional weight, 
so is the depth of the lesser beam to the depth of the 
greater beam. 

Note.—Any other corresponding sides will answer 
the same purpose, for they are all proportioned to one 
another. 

Example. Suppose a beam whose weight is one 
pound, and its length 10 feet, to carry a weight of 
399.5 pounds, required the length of a beam similar 
to the former, of the same matter, so that it shall break 
with its own weight. 

here a=l 

and w=399.5 

then a+2 w=800=1+2X399.5 
d=10 

Then by the last problem it will be 1: 800:10: 8000 
=x for the length of a beam that will break by its own 
weight. 







MECHANICS OP CARPENTRY 


367 


Problem III. The weight and length of a piece of 
timber being given, and additional weight that will 
break it, to find the length of a piece of timber similar 
to the former, so that this last piece of timber shall be 
the strongest possible: 

Put l=the length of the piece given 
w=half its weight, 

W=the weight that will break it; 
x=the length required. 

Then, because the weights that will break similar pieces 
of timber are in proportion to the squares of their 
lengths, 

W x 2 -t- wx 2 

I 2 : x 2 : : W-f-w : -^-= the whole weight that 

breaks the beam; 

and because the weights of similar beams are as the 
cubes of their lengths, or any other corresponding 
sides, • :• 

wx 2 

then l 3 : x 3 : : w : -rr~ the weight of the beam; 

r 


'Wx 2 4- wx 2 wx 3 

consequently -p-less -p- is the weight that breaks 

the beam = a maximum; therefore its fluxion is nothing. 

_ 3wx 2 x . 

‘ that is, 2Wxx + 2 wxx — —j-= nothing. 


nTt7 3wx n £ . 2W-f2w 

2 W + 2w = — j — therefore, x = 1X —g—— 

Hence it appears from the foregoing problems, that 
large timber is weakened in a much greater propor¬ 
tion than small timber, even in similar pieces, there- 









368 


MODERN CARPENTRY 


fore a proper allowance must be made for the weight 
of the pieces, as I shall here show by an 

Example. Suppose a beam 12 feet long, and a foot 
square, whose weight is three hundred pounds, to be 
capable of supporting 20 hundred weight, what weight 
will a beam 20 feet long, 15 inches deep, and 12 thick, 
be able to support? 


12 square inches 

12 


144 

12 


12)1728 


144 


15 

15 


75 

15 


225 

12 


2.0)270.0 


135 


But the weights of both beams are as their solid con¬ 
tents : 

therefore 12 inches square 
12 .. 


144 

144 inches=12 feet long 

576 

576 

144 


20736 solid contents of the 1st beam 














MECHANICS OF CARPENTRY 


369 


15 deep 
12 wide 


180 

240 length in inches 


7200 

360 


43200 solid contents of the 2d beam 


20736 :43200: :3 
3 

-cwt. lb. 

20736) 129600(6 .. 28=the weight of the 2d beam 
124416 


5184 

112 


10368 

5184 

5184 


20736)580608(28 

41472 


16588S 

165888 











370 


MODERN CARPENTRY 


144 ::135: -21.5 by prop. 1. 
21.5 


67.5 

135 

270 


12)2902.5 


12)241.875 


20.25625 

112 


31250 

15625 

15625 


17.50000 

16 


30 

5 


8.0 

21 cwt. 56 lbs. is the weight that will break the first 
beam, and 20 cwt. 17 lb. 8 oz. the weight that will 
break the second beam; deduct out of these half their 
own weight. 

20 : :17 : :8 
3 : :14: :0 half 


17...3..8 

Now 20 cwt. is the additional weight that will break 
the first beam; and 17 cwt. 3 lb. 8 oz. the weight that 












mechanics of carpentry 


37i 


will break the second: in which the reader will ob¬ 
serve, that 10: :3: :8 has a much less proportion to 20, 
than 20 cwt. 17 lb. 8 oz. has to 21::46. From these 
examples, the reader may see that a proper allow¬ 
ance ought to be made for all horizontal beams; that 
is, half the weights of beams ought to be deducted out 
of the whole weight that they will carry, and that will 
give the weight that each piece will bear. 

If several pieces of timber of the same scantling and 
length are applied one above another, and supported 
by props at each end, they will be no stronger than if 
they were laid side by side; or this, which is the same 
thing, the pieces that are applied one above another 
are no stronger than one single piece whose width is 
the width of the several pieces collected into one, and 
its depth the depth of one of the pieces; it is there¬ 
fore useless to cut a piece of timber lengthways, and 
apply the pieces so cut one above another, for these 
pieces are not so strong as before, even if bolted. 

Example. Suppose a girder 16 inches deep, 12 in¬ 
ches thick, the length is immaterial, and let the depth 
be cut lengthways in two equal pieces; then will each 
piece be 8 inches deep, and 12 inches thick. Now, 
according to the rule of proportioning timber, the 
square of 16 inches, that is, the depth before it was 
cut, is 256, and the square of 8 inches is 64; but twice 
64 is only 128, therefore it appears that the two pieces 
applied one above another, are but half the strength 
of the solid piece, because 256 is double 128. 

If a girder be cut lengthways in a perpendicular 
direction, the ends turned contrary, and then bolted 
together, it will be but very little stronger than before 


372 MODERN CARPENTRY 

it was cut; for although the ends being turned give 
to the girder an equal strength throughout, yet 
wherever a bolt is, there it will be weaker, and it is 
very doubtful whether the girder will be any stronger 
for this process of sawing and bolting; and I say this 
from experience. 

If there are two pieces of timber of an equal scant¬ 
ling, the one lying horizontal, and the other in¬ 
clined, the horizontal piece being supported at 
the points e and f, and the inclined piece at c and 
d, perpendicularly over e and f, according to the prin¬ 
ciples of mechanics, these pieces will be equally strong. 
But, to reason a little on this matter, let it be con¬ 
sidered, that although the inclined piece D is longer, 
yet the weight has less effect upon it when placed in 
the middle, than the weight at h has upon the horizontal 
piece C, the weights being the same; it is therefore 
reasonable to conclude, that in these positions the one 
will bear equal to the other. 

The foregoing rules will be found of excellent use 
when timber is wanted to support a great weight; for, 
by knowing the superincumbent weight, the strength 
may be computed to a great degree of exactness, so 
that it shall be able to support the weight required. 
The consequence is as bad when there is too much tim¬ 
ber, as when there is too little, for nothing is more 
requisite than a just proportion throughout the whole 
building, so that the strength of every part shall 
always be in proportion to the stress; for when there 
is more strength given to some pieces than others, it 
encumbers the building, and consequently the foun¬ 
dations are less capable of supporting the superstruc¬ 
ture. 


MECHANICS OF CARPENTRY 


373 


No judicious person, who has the care of construct¬ 
ing buildings, should rely on tables of scantlings, such 
as are commonly in books; for example, in story posts 
the scantlings, according to several authors, are as 
follows: 

For 9 feet high 6 inches square 

12-8 

.15-10 

18-12 

Now, according to this table, the scantlings are in¬ 
creased in position to the height; but there is no pro¬ 
priety in this, for each of these will bear weight in 
proportion to the number 9, 16, 25, and 36, that is, 
in proportion to the square of their heights, 36 being 
4 times 9; therefore the piece that is 18 feet long, will 
bear four times as much weight as that piece which 
is 9 feet long; but the 9 feet piece may have a much 
greater weight to carry than an 18 feet piece, suppose 
double: in this case it must be near 12 inches square 
instead of 6. The same is also to be observed in 
breast-summers, and in floors where they are wanted 
to support a great weight; but in common buildings, 
where there are only customary weights to support, 
the common tables for floors will be near enough for 
practice. 

To conclude the subject, it may be proper to notice 
the following observations which several authors have 
judiciously made, viz.; that in all timber there is 
moisture, wherefore all bearing timber ought to have 
a moderate camber, or roundness on the upper side, 
for till that moisture is dried out, the timber will swag 
with its own weight. 





374 


MODERN CARPENTRY 


But then observe, that it is best to truss girders when 
they are fresh sawn out, for by their drying and 
shrinking, the trusses become more and more tight. 

That all beams or ties be cut, or in framing forced 
to a roundness, such as an inch in twenty feet in 
length, and that principal rafters also be cut or forced 
in framing, as before; because all joists, though ever 
so well framed, by the shrinking of the timber and 
weight of the covering will swag, sometimes so much 
as not only to be visible, but to offend the eye: by 
this precaution the truss will always appear well. 

Likewise observe, that all case-bays, either in floors 
or roofs, do not exceed twelve feet if possible; that is, 
do not let your joints in floors exceed twelve feet, nor 
your purlines in roofs, &c., but rather let their bearing 
be eight, nine, or ten feet. This should be regarded 
in forming the plan. 

Also, in bridging floors, do not place your binding 
or strong joists above three, four, or five feet apart, 
and take care that your bridging of common joists are 
not above twelve or fourteen inches apart, that is, be¬ 
tween one joist and another. 

Also, in fitting down tie-beams upon the wall plates, 
never make your cocking too large, nor yet too near 
the outside of the wall plate, for the grain of the wood 
being cut across in the tie-beam, the piece that remains 
upon its end will be apt to split off, but keeping it 
near the inside will tend to secure it. 

Likewise observe, never to make double tenons for 
bearing uses, such as binding joists, common joists, or 
purlins; for, in the first place, it very much weakens 
whatever you frame it into, and in the second place, 


MECHANICS OF CARPENTRY 


375 


it is a rarity to have a draught to both tenons, that is, 
to draw both joists close; for the pin in passing through 
both tenons, if there is a draught in each, will bend 
so much, that unless it be as tough as wire, it must 
needs break in driving, and consequently do more hurt 
than good. 

Roots will be much stronger if the purlins arc 
notched above the principal rafters, than if they are 
framed into the side of the principals; for by this 
means, when any weight is applied in the middle of 
the purlin, it cannot bend, being confined by the other 
rafters: and if it do, the sides of the other rafters must 
needs bend along with it; consequently it has the 
strength of all the other rafters sideways added to it. 


This volume will be followed by at least three, more, 
on the subject of Carpentry and Joining, one volume 
of which will be devoted to Framing and Heavy Tim¬ 
bered Work. The other two or more volumes will be 
devoted to fine joining and cabinet work, so far as the 
latter art comes within the purview of the joiners’ 
work. 

Each volume will, of course, be illustrated with dia¬ 
grams and working details. The matter will be taken 
from executed works and from every available source. 





INDEX 


A 


An Annular Vault . 

An Ogee Dome . 

Angle Brackets for Core. 

Angle Bracket for Curved and Straight Walls. .. . 

Angle Bracket on a Rake. 

A Frame Splayed Lining. 

A Window with Splayed Soffit. 

Angle Sash Bars ... 

Axle Pulleys . 

A Lantern Light .. 

A Bay Window . 

A Cased Frame Bay Window. 

A Riveted Light . 

A Bull’s-Eye Frame . 

A Segment Headed Frame. 

A Semi-Headed Frame . 

A Solid Frame . 


128 
129 . 
153 
155 
157 
175 
178 
187 
213 
233 
237 
240 
243 
247 
286 

287 

288 


Architrave Grounds.. 294 

Architraves . 295 

A Variety of Panel Doors.299-303 



Bracketing for Cornices . 

Brackets for Re-entrant Angles 

377 


153 

154 


j * 

























378 


INDEX 


Bracket for Curved Cornice. 155 

Boxing Shutters . 191 

Balancing Shutters . 193 

Bay Windows . 216 

Box Shutters . 225 

Barge Boards . 230 

Battered Doors . .. 252 

Brased Doors. 252 

Bead Flush Doors. 255 

Bead and Butt Doors. 256 

Bolection Moulded Doors. 257 

Bead Flush Panels . 259 

Bead Butt Panels.•• 259 

Bolection Mouldings . 261 

Bed-room Doors. 263 

Blind Screwing. 281 

Blind Nailing . 281 

Bending Ornaments . 345 

C 

Cube . 13 

Cube Projection . 44 

Cube and Octagon. 45 

Cone, Sphere, and Cylinder. 57 

Cylinder, Cone and Sphere... 57 

Conic Sections . 72 

Cones from Parabolas. 75 

Cones and Hyperbolas. 76 

Cuneoids . 7^ 

Cylinder Sections . 78 

Cylindrical Rings . 82 

Cylinder and Sphere.... 9Q 
































INDEX 379 

Coverings of Solids. 96 

Covering Prisms . 98 

Covering of Pyramids. 99 

Covering of Right Cylinders. 103 

Covering of Oblique Cylinders. 105 

Covering Semi-Cylindric Surfaces. 106 

Cone Coverings ... 108 

Covering a Segmental Dome. 117 

Covering an Elliptic Dome. 122 

Covering an Ellipsoidal Dome. 123 

Covering an Annular Vault. 128 

Covering an Ogee Dome. 129 

Cutting Oblique Spouts or Timbers. 143 

Cutting Different Pitches. 144 

Ceilings in Wood or Stucco. 146 

Cove Brackets . 153 

Cornice Bracketing . 153 

Corner Bracket on a Rake. 157 

Curb Ribs for Circular Ceiling. 158 

Circular Splayed Work. 176 

Construction of Windows. 201 

Casement Windows . 204 

Cased Pulley Styles. 209 

Circular Head Windows. 220 

Corner Posts and Cross-Rails. 229 

Conservatories . 236 

Chamfered Doors . 256 

Constructive Memoranda . 257 

Common Doors . ^04 

D 

Diagram Showing Solid Geometry. 10 

Dodecahedron . 

































I 

380 INDEX 

Dodecahedron Solids . 51 

Development of Coverings. 07 

Diagram for Oblique Cuts. 113 

Double Semi-Circular Mitre. 100 

Double Curvature Work. 106 

Development of Conical Surfaces. 177 

Diminishing of Mouldings. 180 

Double Tenons . 203 

Double Sunk Rails. 208 

Double Hung Sashes . 214 

Double Bolection Moulded Doors. 257 

Double Margined Doors. 267 

Door and Window Grounds. 279 

Door and Jambs Complete. 298 

Deflection of Beams and Girders. 354 

E 

Ellipsoids . 86 

Ellipsoidal Coverings . 114 

Elliptical Dome Coverings. 120 

Elliptical Niche. 162 

Enlarging of Mouldings. ISO 

Entrance Doors . 263 

External Doors. 264 

External Door Frames. 283 

F 

Finding Projection Points in a Cone. 60 

Finding Point of Projection in Sphere. 64 

Finding an Ellipsoid. 81 

Frustrum of Cones. Ill 






























Index 


38f 


Finding the Stretchout of Curves. 131 

Framed Splayed Circular Soffit. 175 

Folding Shutters . 188 

Franking Sash Bars. 201 

Fixing Window Frames . 221 

Finials . 230 

Framed and Braced Doors. 253 

Framed Paneled Doors. 254 

Framed Grounds . 295 

Forces which Balance Each Other. 310 

G 

Given Points in Planes. 32 

Generating Curves . 34 

Gable Mitres . 140 

Grounds . 226 

Greenhouses and Conservatories. 236 

Grounds for Base. 291 

H 

Horizontal Traces . 24 

Horizontal Projection. 28 

Helix in Cylinder. 62 

Helix in a Cone. 63 

Helical Curve on Sphere. 65 

Hexagonal Pyramid . 67 

How to Cut Oblique Spouts, etc. 143 

Hook Point. 

Hinged Skylights . 227 

Heavy Architraves . 296 




























382 


INDEX 


I 


Intersecting Lines . 

Intersection of Curved Surfaces. .. 

Intersection of Cylinders. 

Intersecting a Sphere and Cylinder 

Intersection of Cones. 

Irregular Mitres . 

Interior Shutters. 

Inclined Forces . 

J 

Joints of Sashes. 

L 


18 

86 

87 

90 

91 
133 
188 
323 


203 


Lamet Curves . 

Large Panels for Ceilings. 

Long Shutters . 

Lifting Shutters . 

Linings . 

Large Sky-Lights . 

Lantern Lights . 

Laying Out a Circular Louvre 

Louvre Boards. 

Louvre Gains . 

Loaded Beams . 


M 

Mitres for Irregular Forms. . 

Mitres on a Rake. 

Mitring Irregular Mouldings 
Mitring Sprung Mouldings. . 


84 

151 

189 

193 

223 

231 

231 

248 

249 

250 
331 


133 

135 

138 

185 


























INDEX 


383 


Mitring Curved and Straight Mouldings. 185 

Mullions . 215 

Moulded and Square Doors.t. 255 

Moulded Grounds.-. 294 

Mechanics of Carpentry. 305 

Methods of Arranging Beams. 344 


N 


Niches . 159 

Niche on Elliptical Plane.. 164 


0 

Octagon Diagram . 

Octagon and Cube. 

Octahedron . 

Octahedron Projection 

Oblique Planes. 

Octagonal Pyramid . 

Oblique Axes . 

Ogee Pyramids. 

Oblique Cuts . 

On Mitres. 

Octagonal Mitre . 

Outside and Inside Linings 
Other Paneled Framing 


13 

44 

46 

47 
57 

69 

70 
85 

143 

159 

163 

209 

278 


P 


Preface . 

Part I. 

Points, Lines and Planes.... 

Planes and Projections. 

Perspective Lines and Traces 


9 

16 

22 

26 




























384 


INDEX 


Projection of Cylinder. 

Pyramid . 

Prisms . 

Pyramids Developed . 

Practical Solutions. 

Projection of Solids. 

Paneled Ceilings, Wood or Stucco 

Parting Strips . 

Pocket Piece . 

•< $ S J 3 3 1 3 J i «. ■* 

Position to Cut Beads. 

Planted Moulded Doors. 

Projections of Doors . 

Plain Jamb Linings . 

Plinth Blocks .. 

Panel Doors in Various Styles. 

Polygon of Forces. 

Parallel Forces . 

Pulleys. 


58 

67 

98 

99 
131 
141 
146 
210 
212 
245 
258 
263 
289 
297 
300 
319 
324 
356 


R 


Right Lines . 36 

Regular Polyhedrons . 96 

Raking Moulding Mitres. 136 

Raking Bracket for Corner. 157 

Ribs for Circular Arch and Ceiling. 158 

Ribs for Octagonal Niche. 163 

Regular Octagonal Niche . 165 

Ribs for Irregular Octagonal Niche.... 166 

Radial Sash Bars. 173 

Raking Mouldings.183 

Raking Sash Bars. 187 

Rolling Shutters . 188 

































INDEX 


385 


Raised Panel Door. 256 

Raised and Flat Door. 256 

Raised, Sunk and Moulded Door. 256 

Raised, Sunk and Moulded. 257 

Reception Room Doors . 263 

Resistant of Two or More Forces. 306 

Reciprocal Diagrams . 332 

Revolving Doors . 2,78 

' . S 

Solid Geometry . 9 

Straight-Sided Solids. 36 

Sphere and Tetrahedron. 55 

Sphere, Cone and Cylinder. 57 

Sections of Solids. 66 

Sections of Cones. 76 

Sections of Cones from Parabolas. 75 

Sections of Cones from Hyperbolas. 76 

Sections of Cuneoids. 78 

Sections of a Cylinder. 78 

Solids of Revolution. 83 

Sf)here and Cylinder . 90 

Semi-Cylindric Coverings . 106 

Spheres and Ellipsoids. 113 

Spherical Niche . 159 

Semi-Circular Niche . 160 

Segmental Niche . 161 

Semi-Elliptical Niche. 163 

Segment Heads . 170 

Staving for Circular Heads.. 170 

Sash Bars for Circular Windows. 173 

Splayed Work Circulars. 176 


































386 INDEX 

Sprung Mouldings . 185 

Shutters. 188 

Sliding Shutters . 193 

Spring Shutters . 197 

Size and Position of Windows. 201 

Sashes . 201 

Sashes Opening Inward. 205 

Sashes Opening Outward. 208 

Sash and Frame Window. 209 

Sashes and Window Frames. 209 

Sashes of Various Kinds. 212 

Single Hung Sashes. 214 

Sliding Sashes . 215 

Shop Windows . 221 

Sliding Shutters . 226 

Sash Frame . 228 

Side-Lights . 233 

Square and Sunk Doors. 254 

Stop Chamfered Doors. 257 

Solid Moulded Doors. 258 

Superior Internal Doors. 264 

Square Shoulder Rail. 271 

Superior Doors . 279 

Secret Fastening. 281 

Solid Door Frames. 283 

Semi-Circular Frame . 288 

Skeleton Frame. 289 

Set of Linings. 289 

Stress Diagrams for Trusses. 337 

Strength of Wooden Beams. 342 

Stiffners of Beams. 342 

Specific Gravity . 358 

Strength of Timber. 359 



































INDEX 


387 


T 

To Determine Length of Oblique Lines. 16 

To Determine Various Angles. 21 

Traces of Intersecting Planes. 25 

Tetrahedron . 37 

Tetrahedron Projection . 40 

To Find Inclination of Two Faces. 49 

To Describe a Helix. 64 

The Cone. 71 

To Describe the Section of a Sphere. 80 

To Describe an Ellipsoid. 81 

To Intersect Two Unequal Cylinders. 88 

To Find Covering of Frustrum of Cone. 112 

To Find the Covering of an Elliptic Dome. 122 

To Find Angle Bracket for Cornice. 154 

To Obtain Soffit Mould. 166 

To Form a Head. 168 

To Obtain a Developed Face Mould. 172 

To Find Mould for a “Cot Bar”. 173 

To Find Mould for Radial Bars. 173 

To Obtain Face Moulds. 176 

To Apply the Moulds. 177 

The Development of Conical Surfaces. 177 

To Diminish a Moulding. 181 

Throating . 205 

To Hang the Sash. 245 

To Find Position to Cut Beads. 245 

The Construction of Doors. 251 

To Set Out Gunstock Stiles. 270 

To Set Out Rail for Gunstock Stile. 271 

The Nature of Force. 306 

The Parallelogram of Forces. 308 

































388 


INDEX 


Triangle of Forces. 3l2 

Transverse Strength of Wooden Beams. 349 

Y 

* ^ * » » 

Vertical and Horizontal Projections. 17 

Vertical Projection . 31 

Vertical Projection of Dodecahedron. 53 

Veneering Soffits. 170 

Venetian Windows . 215 

Vertical Sashes . 215 

W 

Windows Generally . 201 

Window Frames . 204 

Window and Sash-Frames. 210 

Windows with Curved Heads. 219 

Windows for Shops. 221 

Wedges . 221 

Window Linings . 223 

Window Shutters . 225 

Window Grounds . 279 



















NOTICE 

To the many workmen who are purchasing the publication* under the 
authorship of Fred T. Hodgson, and who we feel sure have been benefited 
by his excellent treatises on many Carpentry and Building subjects, we 
desire to inform them that the following list of books have been published 
since 1903, thereby making them strictly up-to-date in every detail. All of 
the newer books bearing the imprint of Frederick J. Drake & Co. are modern 
in every respect and of a purely self-educational character, expressly issued 
for Home Study. 

PRACTICAL USES OF THE STEEL SQUARE, two volumes, over 500 
pages, including 100 perspective views and floor plans of medium- 
priced houses. Cloth, two volumes, price $2.00. Half leather, 
price $3.00. 

MODERN CARPENTRY AND JOINERY, 300 pages, including 50 house 
plans, perspective views and floor plans of medium and low-cosfc 
houses. Cloth, price $1.00. Half leather, price $1.50. 

BUILDERS 1 ARCHITECTURAL DRAWING SELF-TAUGHT, over 350 
pages, including 50 house plans. Cloth, price $2.00. Half leather, 
price $3.00. 

MODERN ESTIMATOR AND CONTRACTORS' GUIDE, for pricing build¬ 
ers’ work, 350 pages, including 60 house plans. Cloth, price $1.50. 
Half leather, price $2.00. 

MODERN LOW-COST AMERICAN HOMES, over 200 pages. Cloth, price 
$1.60. Half leather, price $1.50. 

PRACTICAL UP-TO-DATE HARDWOOD FINISHER, over 300 pages. 
Cloth, price $1.00. Half Leather, price $1.50. 

COMMON SENSE STAIR BUILDING AND HANDRAILING, over 250 
pages, including perspective views and floor plans of 50 medium-priced 
houses. Cloth, price $1.00. Half leather, price $1.50. 

STONEMASONS’ AND BRICKLAYERS’ GUIDE, over 200 pages. Cloth, 
price $1.50. Half leather, price $2.00. 

PRACTICAL WOOD CARVING, over 200 pages. Cloth, price $1.50. Half 
leather, price $2.00. 

Sold by booksellers generally, or sent, all charges paid, upon receipt of 
price, to any address in the world 

FR.EDER.ICK J. DRAKE CO. 

PUBLISHERS OF SELF-EDUCATIONAL BOOKS 

350 352 WABASH AVE.. CHICAGO. ILL. 




A MODERN treatise on Hot Water, Steam and Furnace 
Heating, and Steam and Gas Fitting, which is in¬ 
tended for the use and information of the owners of build¬ 
ings and the mechanics who install the heating plants in 
them. It gives full and concise information with regard 
to Steam Boilers and Water Heaters and Furnaces, Pipe 
Systems for Steam and Hot Water Plants. Radiation, Radi¬ 
ator Valves and connections, Systems of Radiation, Heating 
Surfaces, Pipe and Pipe Fittings, Damper Regulators, Fit¬ 
ters’ Tools, Heating Surface of Pipes. Installing a Heating 
Plant and Specifications. Plans and Elevations of Steam 
and Hot Water Heating Plants are shown and all other sub¬ 
jects in the book are fully illustrated. 

256 pages, 121 illustrations, 12mo, cloth, price, $1.50 

Sold by Booksellers generally or sent postpaid to 
any address upon receipt of price by the Publishers 

FREDERICK J. DRAKE & CO. 

350-352 WABASH AVENUE. CHICAGO. U.S.A. 


























Ql PRACTICAL up-to-date work on Sanitary Plumbing, com- 
prising useful information on the wiping and soldering of 
lead pipe joints and the installation of hot and cold water and 
drainage systems into modern residences, 'including the 
gravity tank supply and cylinder and tank system of water 
heating and the pressure cylinder system of water heating. 
Connections for bath tub. Connections for water closet. 
Connections for laundry tubs. Connections for wash-bowl or 
lavatory. A modern bath room. Bath tubs. Lavatories. 
Closets. Urinals. Laundry tubs. Shower baths. Toilet 
room in office buildings. Sinks. Faucets. Bibb-cocks. Scil- 
pipe fittings. Drainage fittings. Plumber’s tool kit, etc., etc. 
256 pages, 180 illustrations. 

12 Mo. Cloth.$1.50 


Sold by Booksellers generally or sent postpaid to 
any address upon receipt of price by the Publishers 

FREDERICK J. DRAKE ® CO. 

350-352 Wabash Ave., Chicago, U.S. A. 




































IBuiliirr attb 
Qlonlrartor a (Suit)? 

TO CORRECT MEASUREMENTS of areas and 
cubic contents in all matters relating to buildings of any 
kind. Illustrated with numerous diagrams, sketches and 
examples showing how various and intricate measure¬ 
ments should be taken :: :: :: :: :: ” " " :: 

By Fred T. Hodgson, Architect, and W. M. Brown, C.E. and Quantity Surveyor 

® HIS is a real practical book, 
showing how all kinds of 
odd, crooked and difficult meas¬ 
urements may be taken to 
secure correct results. This 
work in no wa} r conflicts with 
any work on estimating as it 
does not give prices, neither 
does it attempt to deal with 
questions of labor or estimate 
how much the execution of cer¬ 
tain works will cost. It simply 
deals with the questions of 
areas and cubic contents of any 
given work and shows how 
their areas and contents may 
readily be obtained, and fur¬ 
nishes for the regular estimator 
the data upon which he can 
base his prices. In fact, the 
work is a great aid and assist¬ 
ant to the regular estimator 
and of inestimable value to the 
general builder and contractor. 


12mo, cloth, 300 pages, fully illustrated, price - $1.50 

Sold by Booksellers generally or sent postpaid to 
any address upon receipt of price by the Publishers 

FREDERICK J. DRAKE & CO. 

350-352 WABASH AVE. *. CHICAGO. U.S.A. 



















































































































































Twentieth Century 
M.achine Shop Practice 


By L. ELLIOTT BROOKES 



Properties of steam, The Indi¬ 
cator, Horsepower and Electricity. 

The latter part of the book gives full and complete information 
upon the following subjects: Measuring devices, Machinists’ tools. 
Shop tools, Machine tools, Boring machines, Boring mills. Drill 
presses, Gear Cutting machines, Grinding Machines, Lathes and Mill¬ 
ing machines. Also auxiliary machine tools, Portable tools, Miscella¬ 
neous tools, Plain and Spiral Indexing machines, Notes on Steel, Gas 
furnaces, Shop talks, Shop kinks, Medical Aid and over Fifty tables. 


The best and latest and most 
practical work published on mod¬ 
ern machine shop practice. This 
book is intended for the practical 
instruction of Machinists, Engin¬ 
eers and others who are interested 
in the use and operation of the 
machinery and machine tools in a 
modern machine shop. The first 
portion of the book is devoted to 
practical examples in Arithmetic, 
Decimal Fractions, Roots of Num¬ 
bers, Algebraic Signs and Symbols, 
Reciprocals and Logarithms of 
Numbers, Practical Geometry and 
and Mensuration. Also Applied 
Mechanics—which includes: The 
lever, The wheel and pinion. The 
pulley, The inclined planes, The 
wedge The, screw and safety valve 
—Specific gravity and the velocity 
of falling bodies—Friction, Belt 
Pulleys and Gear wheels. 


The book is profusely illustrated and shows views of the latest 
machinery and the most up-to-date and improved belt and motor- 
driven machine tools, with full information as to their use and opera¬ 
tion. It has been the object of the author to present the subject 
matter in this work in as simple and not technical manner as is 
possible. 


12mo, cloth, 636 pages, 456 fine illustrations, price, $2.00 


Sold by Booksellers generally, or sent postpaid to 
any address upon receipt of Price by the Publishers 

FREDERICK J. DRAKE & CO. 

350-352 Wabash Ave., CHICAGO, U. S. A. 

















Sheet 

Metal 

Workers ' 1 

Instructor 

By 

Joseph H. Rose 


Profusely Illustrated. 

* [ / HIS work consists of useful information for Sheet Metal 
* Workers in all branches of the industry, and contains 
practical rules for describing the various patterns for sheet 
iron, copper and tin work. Geometrical construction of 
plane figures. Examples of practical pattern drawing. 
Tools and appliances used in sheet metal work. Examples 
of practical sheet metal work. Geometrical construction 
and development of solid figures. Soldering and brazing. 
Tinning. Retinning and galvanizing. Materials used in 
sheet metal work. Useful information. Tables, etc. 

320 Pages, 240 Illustrations 
12 Mo. Cloth, s w . Price, $2.00 

Sold by Booksellers generally, or sent postpaid to 
any address upon receipt of price by the Publishers 

Frederick J. Drake Co. 

350-352 Wabash Ave., CHICAGO. U. S. A. 
































Cements, 
Mortars, 
Plasters 

©Lr\d 

Stuccos 

How to Make and 
How to Use Them 

Fred T. Hodgson 

Architect 

T HIS is another of Mr. Hodgson’s practical works that appeals 
directly to the workman whose business it is to make and apply 
the materials named in the title. As far as it has been possible 
to avoid chemical descriptions of limes, cements and other materials, 
and theories of no value to the workman, such has been done, and 
nothing has been admitted into the pages of the work that does not 
possess a truly practical character. 

Concretes and cements have received special attention, and the 
latest methods of making and using cement building blocks, laying 
cement sidewalks, putting in concrete foundations, making cement 
casts and ornaments, are discussed at length. Plastering and stucco 
work receive a fair share of consideration and the best methods of 
making and using are described in the usual simple manner so 
characteristic of Mr. Hodgson’s style. The book contains a large 
number of illustrations of tools, appliances and methods employed 
in making and applying concretes, cements, mortars, plasters and 
stucco, which will greatly assist in making it easy for the student to 
follaw and understand the text 
300 pages fully illustrated. 

12 Mo. Cloth, . Price, $1.50 

Sold by Booksellers generally or sent postpaid to 
any address upon receipt of price by the Publishers 

Frederick J. Drake Co. 

350-352 Wa-batsh Ave., CHICAGO, U S. A. 


Concretes, 













The AMATEUR 

ARTIST 


CrD 

Or Oil and Water Color 
Painting without the Aid 
of a Teacher :: :: :: :: 

By F. DELAMOTTE 

Cfl The aim of this book is to 
instruct the student in the fund¬ 
amental principles underlying 
those branches of art of which it treats and to teach the application 
of those principles in a clear and concise manner. The knowledge 
it contains is available, alike to the amateur whose only desire it is to 
beautify the home and to pass pleasant hours at agreeable work and 
also to those talented ones who lack the opportunities afforded by art 
schools and teachers who are out of reach. To the latter, this work 
contains elements that will quicken the germ of talent or genius into 
life and send it well on its road to success. CJ This very late and most 
complete work on amateur art gives thorough instructions in nine 
branches of decorative art. Each part is the product of the pen of a 
famous teacher and lecturer who has made that branch his especial 
life study. •]) Unlike other works on the market, it is brought up-to- 
date—no obsolete branches being dragged in, to fill out space. <J Each 
chapter contains a complete list of materials and equipment, and 
instruction enough to develop natural ability to a point where the 
student may continue, independent of further aid, and trusting to his 
own individuality of style. 

200 pages, fully illustrated , price $1.00 

Sold by Bookseller* generally or sent postpaid to 
any address upon receipt of price by the Publishers 

FREDERICK J. DRAKE ©> CO. 

350-352 WABASH AVE. s. CHICAGO. U.S.A. 
































®fie look of Collected 


and 

Engraved 
by F. 

Delamotte 




Large oblong 
octavo, 208 
pages, 100 
designs 
Price, $1.50 


N. B.—We 

guarantee this 
book to be 
the largest and 
best work of 
this kind 
published 


pLAIN and Ornamental, ancient and mediaeval, from the 
Eight to the Twentieth Century, with numerals. In¬ 
cluding German, Old English, Saxon, Italic, Perspective, 
Greek, Hebrew, Court Hand, Engrossing, Tuscan, Riband, 
Gothic, Rustic, and Arabesque, with several Original De¬ 
signs and an Analysis of the Roman and Old English Alpha¬ 
bets, Large, Small, and Numerals, Church Text, Large and 
Small; German Arabesque; Initials for Illumination, Mono¬ 
grams, Crosses, etc., for the use of Architectural and En¬ 
gineering Draughtsmen, Surveyors, Masons, Decorative 
Painters, Lithographers, Engravers, Carvers, etc. 

Sold by Booksellers generally or send postpaid to 
any address upon receipt of price by the Publishers 


iFrpiurtrk 31. Irak? $c (ttnmpatty 

350-352 WABASH AVENUE ::: CHICAGO, U.S.A. 















PRACTICAL BUNGALOWS 
AND COTTAGES FOR 
TOWN AND COUNTRY 



THIS BOOK CONTAINS PERSPECTIVE 
DRAWINGS AND FLOOR PLANS 


OF ONE HUNDRED LOW AND MEDIUM PRICED HOUSES 
RANGING FROM FOUR HUNDRED TO FOUR THOUSAND 
DOLLARS EACH. ALSO TWENTY-FIVE SELECTED 
DESIGNS OF BUNGALOWS FOR SUMMER AND COUNTRY 
HOMES, FURNISHING THE PROSPECTIVE BUILDER 
WITH MANY NEW AND UP-TO-DATE IDEAS AND SUG¬ 
GESTIONS IN MODERN ARCHITECTURE. 

THE HOUSES ADVERTISED IN THIS BOOK ARE EN¬ 
TIRELY DIFFERENT IN STYLE FROM THOSE SHOWN 
IN HODGSON’S LOW COST HOMES. 


12 MO. CLOTH, 200 PAGES, 300 ILLUSTRATIONS 
PRICE, POSTPAID $1.00 


| FREDERICK J. DRAKE & CO. 


350-352 WABASH AVENUE 


CHICAGO 














A WORK of practical information for the use of Owners, Operators and 
Automobile Mechanics, giving full and concise information on all 
questions relating to the construction, care and operation of gasoline and 
electric automobiles, including Road Troubles, Motor Troubles, Carbu¬ 
reter Troubles, Ignition Troubles, Battery 
Troubles, Clutch Troubles, Starting 
Troubles. Pocket size, 4x6%. Over 200 
pages. With numerous tables, useful 
rules and formulas, wiring diagrams and 
over 100 illustrations, by 

L. ELLIOTT BROOKES 

Author of the “Construction of a Gasoline Motor.” 


This work has been written especially 
for the practical information of automo¬ 
bile owners , operators and automobile 
mechanics. Questions will arise, which 
are answered or explained in technical 
books or trade journals, but such works 
are not always at hand, or the information 
given in them not directly applicable to the 
case in hand. 

In presenting this work to readers who may be interested in automo¬ 
biles, it has been the aim to treat the subject-matter therein in as simple 
and non-technical a manner as possible, and yet to give the principles, 
construction and operation of the different devices described, briefly and 
explicitly* and to illustrate them by showing constructions and methods 
used on modern types of American and European cars. 

The perusal of this work for a few minutes when troubles occur, will 
•ften not only save time, money and worry, but give greater confidence in 
the car, with regard to its going qualities on the road, when properly and 
intelligently cared for. 

In conclusion it may be stated that at the present time nearly all auto¬ 
mobile troubles or breakdowns may, in almost every case, be traced to 
the lack of knowledge or carelessness of the owner or operator of the car. 
rather than to the car itself. 16mo. 320 pages, and over 100 illustrations. 

Popular Edition, Full Leather Limp. Price net. $1.50 

Edition d© Luxe, Full Red Morocco, Gold Edges. Price net.. 2.00 



Sent Postpaid to any Address in the Wo rld upon Receipt of Price 

FREDERICK J. DRAKE & CO. 


PUBLISHERS 


350-352 Wabash Avenue, CHICAGO, ILL* 












SEP 6 1906 








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